Find the last two digits of (521)^125 * (125)^521

A sheet of paper contains the numbers 101, 102, 103, ..., 200 written on it. The following process is applied on any two numbers on the sheet.

At each stage, you pick two numbers on the sheet, say a and b, cross these out and replace them with a single number a + b + ab. You keep doing this until only one number is left on the sheet. What is the last number left on the sheet?

OPTIONS

1)200P100 1
2)201P101 1
3)201P100 1
4)201P102 1

This is a very general qs.
Firstly,take three numbers say,1,2...ur result wud be 5...which is 3!-1
take...............................1,2.3...ur result wud be 23...which is 4!-1
Take..............................1,2,3,4..result wud be ...119.which is 5!-1
One can keep on taking numbers and wud find that....for numbers
1,2,3.....n....d result wud be n!-1
Here...it is 101,102,...200...so the result should be 101*102*103....200*201-1...which is equal to 201P100-1...option 3...Chill Maro yr!

sumit99 Says

An engineer undertakes a project to build a road 15km long in 300 days and employs 45 men for the purpose.after 100 days,he finds only 2.5 km of the road has been completed.find the number of extra men he must employ to finish the work in time.

Finally it's given that working 100 days they will do 2.5 km road. Working at same rate to complete remaining 12.5 km they will need 500 more days.But we have only 200 days.So time has become 2/5=>strength should become 5/2 times the original strength = 5/2 * 45 = 112.5

Thus we need 113 men to finish the work in time =>68 men extra are required

sumit99 Says

An engineer undertakes a project to build a road 15km long in 300 days and employs 45 men for the purpose.after 100 days,he finds only 2.5 km of the road has been completed.find the number of extra men he must employ to finish the work in time.

Problem type:
Input=Output
Let the expected number of hours a laborer puts=x hr.
Therefore,Output Hours=x*45*300hrs.
The actual number of hours a laborer puts=y hrs.
Thus Input(100 days)=y*45*100
Thus, x=2y
y*45*100+y*(45+m)*200=x*45*300
y*(4500+9000+200m)=2y*45*300
13500+200m=27000
m=13500/200=67.5=68
I know thoda bada hai,bt rocksolid hai

1) My take is 49.

2) Bhai 2 aur 3 ki power kaise nikaalu?? 5 ki toh 105 aayegi. :/
Kuch bann sa ni ra.

Coming to 2...

AX2XB where A could 5 or 5^2 and B could be 3^0,3^1,3^2,3^3,3^4
Thus in the form AX2XB, we will have 2 * 5 =10 numbers and in all these numbers we have only 2 and hence in the product we will have 2^10

AX2^2XB, we will have 10 numbers and in the product we will have (2^2)^10 = 2^20

..........

..........

AX2^6XB we will have 2^60

Thus in the entire product 2^(10+20+...+60) = 2^210


Similarly AXBX3, we will have 14 numbers and the product will have 3^14
AXBX3^2, we will have 14 numbers and the product will have 3^28
AXBX3^3, we will have 14 numbers and the product will have 3^42
AXBX3^4, we will have 14 numbers and the product will have 3^56

Thus in the entire product 3^14(1+2+3+4) = 3^140

sumeet1489 Says

What is the product of all factors of the number N = (6^4)*(10^2), which are divisible by 5?

N= 2^6 * 3^4 * 5^2

Product of all factors = N^ * sqrt N

or, N^(7*5*3-1)/2 * 2^3*3^2*5= N^52 * 2^3*3^2*5 = 2^315*3^210*5^105

Now prduct of all factors which are not multiples of 5

R= 2^6 * 3^4

product of all factors of R= R^(7*5 -1)/2 * 2^3* 3^2

=> 2^105 * 3^70

=> PRODUCT OF FACTORS WHICH ARE MULTIPLES OF 5 =( PRODUCT OF ALL FACTORS) / (PRDUCT OF FACTORS WHICH ARE NOT MULTIPLES OF 5)

=> 2^315 * 3^210 * 5^105 / 2^105* 3^70 = 2^210 * 3^140 * 5^105 = ANSWER

1) My take is 49.

9 large boxes.
15 medium boxes in 3 large boxes. Hence, 6 large boxes are empty.
Now, 25 small boxes in 5 medium boxes. Hence,10 medium boxes are empty.
=> Total boxes= 9+15+25 = 49.

2) Bhai 2 aur 3 ki power kaise nikaalu?? 5 ki toh 105 aayegi. :/
Kuch bann sa ni ra.

pehla sahi hai
aur second question mein options diye hue hain jisme se ek hi main 5^105 hai
but I was hoping ki 2 aur 3 ka bhi tareeka mil jaye to concept kuchh clear ho jayega
let's see what others come up with
vaise OA is 2^140 3^102 5^210

PS: big guns ne to mere post se pehle hi method post kar diya :clap::clap::clap:

sumeet1489 Says

What is the product of all factors of the number N = (6^4)*(10^2), which are divisible by 5?

That should be 5^105 * 2^210 * 3^140

N = 2^6 * 3^4 * 5^2

Now, we are looking for factors of N which are divisible by 5.

Exactly choosing one 5 I will have 7 * 5 = 35 numbers (since there are 6 2's, remember i can even leave out 2, I can take 6+1 = 7 chances with 2 and similarly with 3 i can take 5 chances)
When I multiply all these numbers the number of 5's will be 5^35

By choosing both the 5's i.e 5^2 I will have 7*5 = 35 numbers.
When I multiple all these numbers the number of 5's will be (5^2)^35 = 5^70

Thus in the entire product, the number of 5's will 5^35*5^70 = 5^105

Extend the same logic for 2 & 3 as well

The total product will be 5^105 * 2^210 * 3^140

sumeet1489 Says

What is the product of all factors of the number N = (6^4)*(10^2), which are divisible by 5?

It should be 30*(3^5 - 1)(2^7 - 1)

Guess,i found the summation...

editing for the product

1) Richa has three types of boxes viz. large, medium and small. She plays a game in which she placed 9 large boxes on the table. She puts 5 medium boxes each, in a few of the large boxes then she puts 5 small boxes each, in few of the medium boxes. If the number of boxes that have been left empty in the game is 41, then how many boxes were used in the game by Richa?

2)What is the product of all factors of the number N = (6^4)*(10^2), which are divisible by 5?

1)49 hai kya ?

1) Richa has three types of boxes viz. large, medium and small. She plays a game in which she placed 9 large boxes on the table. She puts 5 medium boxes each, in a few of the large boxes then she puts 5 small boxes each, in few of the medium boxes. If the number of boxes that have been left empty in the game is 41, then how many boxes were used in the game by Richa?

2)What is the product of all factors of the number N = (6^4)*(10^2), which are divisible by 5?

1) My take is 49.

9 large boxes.
15 medium boxes in 3 large boxes. Hence, 6 large boxes are empty.
Now, 25 small boxes in 5 medium boxes. Hence,10 medium boxes are empty.
=> Total boxes= 9+15+25 = 49.

2) Bhai 2 aur 3 ki power kaise nikaalu?? 5 ki toh 105 aayegi. :/
Kuch bann sa ni ra.

1) Richa has three types of boxes viz. large, medium and small. She plays a game in which she placed 9 large boxes on the table. She puts 5 medium boxes each, in a few of the large boxes then she puts 5 small boxes each, in few of the medium boxes. If the number of boxes that have been left empty in the game is 41, then how many boxes were used in the game by Richa?

2)What is the product of all factors of the number N = (6^4)*(10^2), which are divisible by 5?

napster11 Says

Q. A number N when divided by a divisor D gives a remainder 52. The number 5N when divided by D gives a remainder 4. How many values of D are possible?

N=KD+52;
5N=PD+4;
N=PD+4/5=KD+52;

D=256;

NOW if we represent 256 as multiplication of 2 numbers and as in one case remainder is 52 then number must be greater than 52..

in 256 's factors 64 is the 1st factor which is greater than 52...

so case1:64*4
case 2:128*2;
case 3:256*1;

so 3 values possible...

find the greatest power of 140, which can exactly divide product of 101*104*107*110*.... (upto 70 terms)
a)1 b)11 c)12 d)13 e)none

method please...

the number of 7 in the product < the number of 5 in the product

7a = 98 + 3b

From this,we can say that b always takes multiple of 7

so,7<= b <= 70,total value is 10

but in one case which is 49 we have 2 sevens,and 98 also multiple of 49

so,Total numbers = 10+1 = 11

rhinorocks Says

mani bhai 1st line thoda elaborate karo.

98 + 3b = 7a or 98 + 3b = 5c

Ideally,we shoulda calculated both a and b,and the lower valve among them has to be considered...

but 95 + 3(b+1) would be divisible for every time when b takes every time 5x+4,and we could conclude c > a...

so,we directly calculated the number of 7...

napster11 Says

Q. A number N when divided by a divisor D gives a remainder 52. The number 5N when divided by D gives a remainder 4. How many values of D are possible?

My take is 3.

N = xD + 52.
5N = yD + 260.
5N = yD + 256 + 4.
So 256 and its factors can be the answer but as remainder given with D is 52 then any factor less then that cant be considered.
Hence D can only be 256,128,64.

What is the product of all factors of the number N = (6^4)*(10^2), which are divisible by 5?

napster11 Says

Q. A number N when divided by a divisor D gives a remainder 52. The number 5N when divided by D gives a remainder 4. How many values of D are possible?

N leaves a remainder 52 (So D>52)
5N must leave a remainder 5*52 = 260
But it is given that the remainder = 4
Hence,the maximum possible value of D = 256 = 2^8
All possible values of D will be nothing but factors of 2^8. Also they must be > 52
Hence D can take 2^6,2^7,2^8. Thus 3 values

find the greatest power of 140, which can exactly divide product of 101*104*107*110*.... (upto 70 terms)
a)1 b)11 c)12 d)13 e)none

method please...

The 70th term of the product = 101+69*3 = 308
We have 140 = 4*5*7
Highest power of 140 that will exactly divide the product will be nothing but the least value among the highest powers of 4 5 and 7 in the given product.
It is evident from the product that the highest power of 7 in the given product will be least.
Let's calculate the highest power of 7 in the given product

Figure out the multiples of 7 in the given product
Every number is of the form 101+3k and k takes values from 0 -69
101+3k divided by 7, the remainder will be of the form 3+3k i.e. 3(k+1)
Thus 101+3k will be divisible by 7 if 3(k+1) is divisible by 7i.e k+1 must be divisible by 7.
All possible of values of k are 6,13,...69 a total of 10 values

Figure out the multiples of 49 in the given product
101+3k when divided by 49, the remainder will be of the form 3(k+1)
For 3(k+1) to be divisible by 49, only possible value of k is 48

Thus the given product has 10 numbers divisible by 7 and one number divisible by 49

Thus the highest power of 7 in the given product is 11

So 11 must be the answer

Q. A number N when divided by a divisor D gives a remainder 52. The number 5N when divided by D gives a remainder 4. How many values of D are possible?

the number of 7 in the product < the number of 5 in the product

7a = 98 + 3b

From this,we can say that b always takes multiple of 7

so,7<= b <= 70,total value is 10

but in one case which is 49 we have 2 sevens,and 98 also multiple of 49

so,Total numbers = 10+1 = 11

mani bhai 1st line thoda elaborate karo..

find the greatest power of 140, which can exactly divide product of 101*104*107*110*.... (upto 70 terms)
a)1 b)11 c)12 d)13 e)none

method please...

the number of 7 in the product < the number of 5 in the product

7a = 98 + 3b

From this,we can say that b always takes multiple of 7

so,7<= b <= 70,total value is 10

but in one case which is 49 we have 2 sevens,and 98 also multiple of 49

so,Total numbers = 10+1 = 11

find the greatest power of 140, which can exactly divide product of 101*104*107*110*.... (upto 70 terms)
a)1 b)11 c)12 d)13 e)none

method please...

bhai 1 hai kya..?
ye 101*104*-----140*143....308 aisa series hai...
so 140^1 will do it..


ye maine lowest ka nikal diya.....

find the greatest power of 140, which can exactly divide product of 101*104*107*110*.... (upto 70 terms)
a)1 b)11 c)12 d)13 e)none

method please...

Lowest number is 840-2 =838.

next number will be 838 + 840k= 9t

lhs has to be a multiple of 9

=> 1 + 3*k

now, k on dividing by 9 will leave remainder of 0,1,2,3,4,5,6,7,8

but as we can see , for no such value f k we will get a multiple of 9.

so my take none.

perfect allan bhai.. i think ur soln is the right one..

find the least difference btw 2 numbers which when divided by 5,6,7 and 8 leave remainder 3,4,5 and 6 but when divided by 9 leave no remainder..
a)2520 b) 5040 c)7560 d)15120 e) none

mere khayal se answer is 2520.. baaki aap log batao aapka kya aa rha hain..

and method please...!!!

Lowest number is 840-2 =838.

next number will be 838 + 840k= 9t

lhs has to be a multiple of 9

=> 1 + 3*k

now, k on dividing by 9 will leave remainder of 0,1,2,3,4,5,6,7,8

but as we can see , for no such value f k we will get a multiple of 9.

so my take none.

ek aur prashna... seems simple.. but me getting confused..;(

Z is family of numbers which are 4 more than multiple of 21 and 10 more than multiple of 35. By which largest number any member of Z can be divided giving same remainder every time..
a)7 b)35 c)105 d)210 e)none

i think answer is 105.. someone please confirm.. also method please...

i am getting the nos as 225K
so according to me the ans is none.......
==================================
P.S==Guess i missed something

ek aur prashna... seems simple.. but me getting confused..;(

Z is family of numbers which are 4 more than multiple of 21 and 10 more than multiple of 35. By which largest number any member of Z can be divided giving same remainder every time..
a)7 b)35 c)105 d)210 e)none

i think answer is 105.. someone please confirm.. also method please...

find the least difference btw 2 numbers which when divided by 5,6,7 and 8 leave remainder 3,4,5 and 6 but when divided by 9 leave no remainder..
a)2520 b) 5040 c)7560 d)15120 e) none

mere khayal se answer is 2520.. baaki aap log batao aapka kya aa rha hain..

and method please...!!!

None i guess

840a + 838 = 9b

=> 837a + 837 + 3a + 1 = 9b

As it's impossible that 3a + 1 = 9b

so,i ll go with none

guess i'm making some silly mistake otherwise...

find the least difference btw 2 numbers which when divided by 5,6,7 and 8 leave remainder 3,4,5 and 6 but when divided by 9 leave no remainder..
a)2520 b) 5040 c)7560 d)15120 e) none

mere khayal se answer is 2520.. baaki aap log batao aapka kya aa rha hain..

and method please...!!!

bhaiyon..m getting 150..don kno its correct or not..

Approach:-lcm of 12,18,9 is 36..
now 3 digit numbers for 36 will start from 108 and last one is 972...

now from 108-144 dere will be 6 numbers that will satisfy the condition..

as 109,110,112,113,115,116...now 117 is divisible by 9...

so in b/w 2 numbers 6 numbers satisfy the given condition..

now in total we have 25 terms from 108-972 so 24*6=144 terms..
but 6 terms will add again from 973-980..as 981 will again be diivisible by 9..
so total 144+6=150

is ithis approach correct?

rayus Says

na re.. the question states that num shud give same remainder when divided by 18, 12 and 9 and no remainder when divided by 3 ...

haan bhai...maine wrong question samajh liya.......:banghead::banghead:

rhinorocks Says

rayus bhai ek baat batao..3 se bhi same remainder hona chahiye?

na re.. the question states that num shud give same remainder when divided by 18, 12 and 9 and no remainder when divided by 3 ...

guys.. my approach...

lcm of 18, 12 and 9 = 36
multples of 36 from 100 to 999 are 108 to 972,, total of 25 multiples...

now for each multiple.. we will get 2 numbers satisfying all the condition..

for ex the number will be of form multiple +3k

so for 108.. 108+3 and 108+6 ... iske aage remainders diff ho jaayenge..
similarly 144 +3 and 144 +6.. and so on for all the 25 multiples...

so total numbers we need it 25*2=50...

batao koi yeh thik hain ya nahin...

Bhai, I have edited my post.do check it I by mistake took 108 as the lcm.

find the least difference btw 2 numbers which when divided by 5,6,7 and 8 leave remainder 3,4,5 and 6 but when divided by 9 leave no remainder..
a)2520 b) 5040 c)7560 d)15120 e) none

mere khayal se answer is 2520.. baaki aap log batao aapka kya aa rha hain.. :)

and method please...!!!

guys.. my approach...

lcm of 18, 12 and 9 = 36
multples of 36 from 100 to 999 are 108 to 972,, total of 25 multiples...

now for each multiple.. we will get 2 numbers satisfying all the condition..

for ex the number will be of form multiple +3k

so for 108.. 108+3 and 108+6 ... iske aage remainders diff ho jaayenge..
similarly 144 +3 and 144 +6.. and so on for all the 25 multiples...

so total numbers we need it 25*2=50...

batao koi yeh thik hain ya nahin...

rayus bhai ek baat batao..3 se bhi same remainder hona chahiye?

bhaiyon..m getting 150..don kno its correct or not..

Approach:-lcm of 12,18,9 is 36..
now 3 digit numbers for 36 will start from 108 and last one is 972...

now from 108-144 dere will be 6 numbers that will satisfy the condition..

as 109,110,112,113,115,116...now 117 is divisible by 9...

so in b/w 2 numbers 6 numbers satisfy the given condition..

now in total we have 25 terms from 108-972 so 24*6=144 terms..
but 6 terms will add again from 973-980..as 981 will again be diivisible by 9..
so total 144+6=150

is ithis approach correct?

bhai jan.. the numbers have to multiples of 3.. as they shudnt leave any remainder when divided by three...

N=108k + 3;

total nos =9

N=108k + 6

total nos = 8

hence total = 9+8=17


mani bhai, gud evening, I almst forgot the direct formula. can u just post it.

P:S bdw u have the same answer as mine.

bhaiyon..m getting 150..don kno its correct or not..

Approach:-lcm of 12,18,9 is 36..
now 3 digit numbers for 36 will start from 108 and last one is 972...

now from 108-144 dere will be 6 numbers that will satisfy the condition..

as 109,110,112,113,115,116...now 117 is divisible by 9...

so in b/w 2 numbers 6 numbers satisfy the given condition..
now in total we have 25 terms from 108-972 so 24*6=144 terms..
but 6 terms will add again from 973-980..as 981 will again be diivisible by 9..
so total 144+6=150

is ithis approach correct?

guys.. my approach...

lcm of 18, 12 and 9 = 36
multples of 36 from 100 to 999 are 108 to 972,, total of 25 multiples...

now for each multiple.. we will get 2 numbers satisfying all the condition..

for ex the number will be of form multiple +3k

so for 108.. 108+3 and 108+6 ... iske aage remainders diff ho jaayenge..
similarly 144 +3 and 144 +6.. and so on for all the 25 multiples...

so total numbers we need it 25*2=50...

batao koi yeh thik hain ya nahin...

eine frage freunde... (a question friends.. :))

how many 3 digit numbers leave same remainder when divided by 18,12 and 9 but no remainder by 3...
a)25 b)50 c)75 d)100 e) none.

i think its 50.. lemeknow if m rite or not... and as always... method please...

bhaiyon..m getting 150..don kno its correct or not..

Approach:-lcm of 12,18,9 is 36..
now 3 digit numbers for 36 will start from 108 and last one is 972...

now from 108-144 dere will be 6 numbers that will satisfy the condition..

as 109,110,112,113,115,116...now 117 is divisible by 9...

so in b/w 2 numbers 6 numbers satisfy the given condition..

now in total we have 25 terms from 108-972 so 24*6=144 terms..
but 6 terms will add again from 973-980..as 981 will again be diivisible by 9..
so total 144+6=150

is ithis approach correct?

---------------------
this is what i get for reading the question wrong !!

N=108k + 3;

total nos =9

N=108k + 6

total nos = 8

N=108k + 9

total nos =8

hence total = 9+8+8=25


mani bhai, gud evening, I almst forgot the direct formula. can u just post it.

P:S bdw u have the same answer as mine.

The formula is basically derived from the fact that how much each digit with its relative position in the number contributing to the total sum..

Having said that,5 in its unit position would contribute 5*(5*5*5*5) to the sum..

Likewise for 5 in 5th position would contribute (5*5*5*1)*5*10000

if it's in 4th position,it'd contribute (5*5*5*1)*5*1000

so,we have to do it till 2nd position ie (5*5*5*1)*5*10

Similarly for the digits - 1,2,3,and 4..that's it

P.S:Allan,Good Evening,

P.S.S:It's better know the logic than the formula...

nitish22691 Says

how can u take 9???????:o:o

yeah, mistake....got it.

N=108k + 3;

total nos =9

N=108k + 6

total nos = 8

N=108k + 9

total nos =8

hence total = 9+8+8=25


mani bhai, gud evening, I almst forgot the direct formula. can u just post it.

P:S bdw u have the same answer as mine.

how can u take 9???????:o:o

snipped...........

eine frage freunde... (a question friends.. :))

how many 3 digit numbers leave same remainder when divided by 18,12 and 9 but no remainder by 3...
a)25 b)50 c)75 d)100 e) none.

i think its 50.. lemeknow if m rite or not... and as always... method please...

Mistake. I took 108 as the lcm.

first 3 digit number leaving same remander will be 108 +3 =111.'

next number will be 111+ lcm(18,12,9) = 111+36=147, and so on..

all such nos will satisfy

hence 111+ (n-1)36 <1000

=>n=25

smallest n of the frm 108 +6 will also satisfy

hence next no will be 114 + lcm(12,18,9)=114+36 =150

and so n

hence 114+ 36*(n-1) <1000

=> n=25

hence total nos willbe 25+25=50

mani0303 Says

My take:5^3*150*1111 + 5^5

mani bhai, gud evening, I almst forgot the direct formula. can u just post it.

P:S bdw u have the same answer as mine.

Find the Sum OF All 5 digit No's ending in 5 (Repetition Is allowed) using the digit 1,2,3,4,5.

My take:5^3*150*1111 + 5^5

Find the Sum OF All 5 digit No's ending in 5 (Repetition Is allowed) using the digit 1,2,3,4,5.

Total nos= 5^4=625 nos.

now each of 1,2,3,4,5 will appear equal no of times at the tens , hundreds...etc places.

=> each digit appears 125 times.

sum of all digits at the tens place = (1+2+3+4+5)125 = 1875

similarly at the hundreds etc places.

sum of all nos.= 10000*1875 + 1000*1875 + 100*1875 + 10*1875+ 625*5

Total sum = 20834375. .wats the OA?

eine frage freunde... (a question friends.. :))

how many 3 digit numbers leave same remainder when divided by 18,12 and 9 but no remainder by 3...
a)25 b)50 c)75 d)100 e) none.

i think its 50.. lemeknow if m rite or not... and as always... method please...

Hi all,

A and B start their journey from Delhi to Agra. B overtakes A at 10 a.m. and
reaches Agra at 1 p.m. On his way back, he meets A at 2 p.m. When will A reach Agra?
A) 3 p.m.
B) 3:30 p.m.
C) 3:50 p.m.
D) 4 p.m.

What is the right approach to solve the above problem ?

Thanks in advance..

answer is 4pm
let dist from point where they are @ 10am to agra be x
at 2 pm when a and b meet then the total distance covered by both of them = 2x

4a+4b=2x

now b reaches agra @ 1pm.. so b=x/3
from above 2 eqns 2a=b

thus a will take double time as compared to b.. 6 hrs from 10 am ... therefore 4pm..

Faruq Says

In an Examination 52% candidate failed in englosh and 42 % failed in MAth. if 17% failed in both subjects. then what percentage is pass in both subjects

failed nly in ENG -> 35%

failed only in math -> 25%

=>passed in both subject= 100%-failed in atleast 1 subject =100% -(25%+35% +17%) =23%

Billi has Six sons namely A,B,C,D,E,F (not in order of their ages) . there are two sets of twins . they are neither the youngest nor the eldest
A & C have only one elder brother D
E is younger to A, but elder to F
The youngest of the sibbling is ????????????????

the twins are AC , and BE.

Hence youngest is F.

Hi all,

A and B start their journey from Delhi to Agra. B overtakes A at 10 a.m. and
reaches Agra at 1 p.m. On his way back, he meets A at 2 p.m. When will A reach Agra?
A) 3 p.m.
B) 3:30 p.m.
C) 3:50 p.m.
D) 4 p.m.

What is the right approach to solve the above problem ?

Thanks in advance..

Q In the diagram shown above, we have twenty-seven 1 by 1 cubes. Each face of every cube is marked with a natural number so that two opposite faces (top and bottom, front and back, left and right) are always marked with an even number and an odd number where the even number is twice that of the odd number. The twenty-seven cubes are put together to form one 3 by 3 cube as shown. When two cubes are placed face-to-face, adjoining faces are always marked with an odd number and an even number where the even number is one greater than the odd number. What is the sum of all of the numbers on all of the faces of all the 1 by 1 cubes?
OPTIONS

1)1379

2)1380

3)1377

4)None

upudit bhai ek bat bta jara...yar in ques...12121212....300 times remainder with 99
mene pahle 9 se remainder dekho =0 aya
11 se dekha to 4 araha hai ...
phir chinese lgayi
9x+11y=0
x=5
y=-4
jise 180 araha hai matlab remainder -18 yar bt ans is 18 help where m going wrong

bhai hogya 11 se remainder 7 aega lag gayi galti pta ..

In how many ways can a three digit code number be formed using the digits 0 to 9 without repetion of digit such that it has only one pair of consecutive digits

sol:

let the code be abc here 0<=a,b,c<=9

case1: when ab is consecutive
a can take 9 values 0 to 8 can't take 9 for this b will 10 out of range
c can take 8 values becoz repetition is not allowed

total values= 9*8=72

similarly when ac & bc are consecutive

total values =72*3=216

upudit bhai ek bat bta jara...yar in ques...12121212....300 times remainder with 99
mene pahle 9 se remainder dekho =0 aya
11 se dekha to 4 araha hai ...
phir chinese lgayi
9x+11y=0
x=5
y=-4
jise 180 araha hai matlab remainder -18 yar bt ans is 18 help where m going wrong

Find the Sum OF All 5 digit No's ending in 5 (Repetition Is allowed) using the digit 1,2,3,4,5.

1)
_ _ _ _ 5


now sum of digit at tens place 1+2+3+4+5=15
rest 100th,1000th,10000th place can arrange in 5*5*5=125 ways
so total sum at tens place = 125*15

similarly for 100th,1000th & 10000th place 125*15

sum
=125*15 (10^4+10^3+10^2+10)+625*5
=20834375

Q Following inequalities represent length of sides of ABC,

a 𕒶

|b 𕒷

c^2 2c 24<0

If a, b and c are integers, what is the difference between the minimum possible and maximum possible perimeter of the ABC?
OPTIONS


1)14

2)11

3)10

4)12

5)13

Find the Sum OF All 5 digit No's ending in 5 (Repetition Is allowed) using the digit 1,2,3,4,5.

In an Examination 52% candidate failed in englosh and 42 % failed in MAth. if 17% failed in both subjects. then what percentage is pass in both subjects

In how many Ways can 7 identical rings be worn in 10 differ. fingers if exactly 5 fingers are to be occupeid?

If a year has 360 days, and all months have 30 days then what is the probability that your birthday falls on monday and that is even day of an even month. if january 1 is a monday

Billi has Six sons namely A,B,C,D,E,F (not in order of their ages) . there are two sets of twins . they are neither the youngest nor the eldest
A & C have only one elder brother D
E is younger to A, but elder to F
The youngest of the sibbling is ????????????????


In how many ways can a three digit code number be formed using the digits 0 to 9 without repetion of digit such that it has only one pair of consecutive digits

A sheet of paper contains the numbers 101, 102, 103, ..., 200 written on it. The following process is applied on any two numbers on the sheet.

At each stage, you pick two numbers on the sheet, say a and b, cross these out and replace them with a single number a + b + ab. You keep doing this until only one number is left on the sheet. What is the last number left on the sheet?

OPTIONS

1)200P100 1
2)201P101 1
3)201P100 1
4)201P102 1

a+b+ab = (a+1)(b+1)-1
We can interpret this as follows
Say I have 4 mangoes (distinguishable) and 5 oranges (distinguishable). In how many ways can i choose at least one fruit.
I can choose any of the 4 mangoes or leave it out. So with mangoes i can take (4+1) chances.Similarly with oranges (5+1) chances.
Total chances will be (4+1)(5+1)
But I cannot leave both the fruits as I have to choose at least one of them.
Hence total number of chances = (4+1)(5+1) - 1
We can generalize this for any number say a,b,c,d
The it will be (a+1)(b+1)(c+1)(d+1) - 1
Coming to our question the final number will (101+1)(102+1)...(200+1) - 1
= 102*103*...*201 - 1
=201P100 - 1

i think u r getting it wrong here

accoding to you ans is 201!/101!==201P100-1 :cheers:

why? I tried it with 2 3 4getting 59means 5!/2! -1 =60 -1that means n+1/ innitital value so here 200+1! / 101! -1201p101 matlab yahi hota hai na 200+1! / 101! me confuse rahti hu isme

My take option 2.
take 1 2 3
getting 23
means 4!/1! -1

so here 201P101 -1

to confirm 100 or 101 try with 2 3 4

i think u r getting it wrong here

accoding to you ans is 201!/101!==201P100-1

A sheet of paper contains the numbers 101, 102, 103, ..., 200 written on it. The following process is applied on any two numbers on the sheet.

At each stage, you pick two numbers on the sheet, say a and b, cross these out and replace them with a single number a + b + ab. You keep doing this until only one number is left on the sheet. What is the last number left on the sheet?

OPTIONS

1) 200P100 1
2) 201P101 1
3) 201P100 1
4) 201P102 1

My take option 2.
take 1 2 3
getting 23
means 4!/1! -1


so here 201!/101! -1

201p100 = 201! / (201-100)! = 201!/101!
to confirm 100 or 101 try with 2 3 4

so option 3