hey cn any1 plz explain the shortest method for solving the following type of ques???
find the last TWO digit of 7^2008
i usually divide it by 100 but the process takes considerably long time,,,,
hey cn any1 plz explain the shortest method for solving the following type of ques???
find the last TWO digit of 7^2008
i usually divide it by 100 but the process takes considerably long time,,,,
7^2008= (7^4)^502=(2401)^502
The units digit will be 1 and for the tens digit, just multiply the tens digit of the number (here, 0) with the last digit of the exponent (2), which will give you 0 => last 2 digits will be 01.
hey cn any1 plz explain the shortest method for solving the following type of ques???
find the last TWO digit of 7^2008
i usually divide it by 100 but the process takes considerably long time,,,,
Hi Apprentice,
Pls see that we have cyclic order for the unit digits for power of a number
I mean 7^1=07 , 7^2=49,7^3=343,7^4=2401
7^5=16807, 7^6=117649, 7^7=823543, 7^8=5764801
So we can cleraly see that unit digit is repeated in cyclic order....
Now also pls observe last two digits of each number, you can see they are too cyclic....
i.e. 7^2=49 and 7^6=117649
also 7^4=2401 and 7^8=5764801
so we can conclude any power of 7 which is multiple of 4 ends wid 01....
so answer to 7^2008 divided by 100 is 01.
hey cn any1 plz explain the shortest method for solving the following type of ques???
find the last TWO digit of 7^2008
i usually divide it by 100 but the process takes considerably long time,,,,
7^1 =07, ends in 07
7^2 =49, ends in 49
7^3 = 343, ends in 43..
7^4 will end in 01
7^5 will end in 07...and so on.
7^4k will end in 01.
2008=4k, So, last two digits are 01.
7^2008= (7^4)^502=(2401)^502
The units digit will be 1 and for the tens digit, just multiply the tens digit of the number (here, 0) with the last digit of the exponent (2), which will give you 0 => last 2 digits will be 01.
Liked your method.:cheerio:
What if we are asked for last three or last four digits? I do get the part where we find unit digit and tens digit. But for finding hundred digit will we multiply the last (two) digit with 0 or 01 or any other? I mean how to proceed then?
Let me know if I'm unclear.
Thanks!
AT
7^2008= (7^4)^502=(2401)^502
The units digit will be 1 and for the tens digit, just multiply the tens digit of the number (here, 0) with the last digit of the exponent (2), which will give you 0 => last 2 digits will be 01.
hey,this goes down well with an odd no in ques(7 here)
suppose it had been 2^134,then hw shud i approach?????
nd also abt the tens digit,r u 100% sure abt the method u hv provided??
hey,this goes down well with an odd no in ques(7 here)
suppose it had been 2^134,then hw shud i approach?????
nd also abt the tens digit,r u 100% sure abt the method u hv provided??
This method works fine for all the number with 1 at the units digit place. You can easily bring a number ending with 3, 7 and 9 to end with 1.
For the numbers ending with 2,4,6 and 8 you can follow this approach.
2^134= *(2^4)= (1024^13)*(16)=> 24*16=384
Hence the number will end with 84.
hey,another problem
wat is this base system??
i came across this ques in LOD2 of no system
wen yuvraj hit 6 sixes in an over,geoff boycott got excited and said that yuvraj has scored 210 runs in an over...,wat base system is he using??????
hey,another problem
wat is this base system??
i came across this ques in LOD2 of no system
wen yuvraj hit 6 sixes in an over,geoff boycott got excited and said that yuvraj has scored 210 runs in an over...,wat base system is he using??????
Base system is 4... Trial n error method!!
hey,another problem
wat is this base system??
i came across this ques in LOD2 of no system
wen yuvraj hit 6 sixes in an over,geoff boycott got excited and said that yuvraj has scored 210 runs in an over...,wat base system is he using??????
Is it 8????
hey,another problem
wat is this base system??
i came across this ques in LOD2 of no system
wen yuvraj hit 6 sixes in an over,geoff boycott got excited and said that yuvraj has scored 210 runs in an over...,wat base system is he using??????
Answer should be base 4.
36 is in decimal. Let the new base be a.
So, 36 = 2*a^2 + 1*a^1 + 0*a^0
Solve the eqaution, a = 4
hey,another problem
wat is this base system??
i came across this ques in LOD2 of no system
wen yuvraj hit 6 sixes in an over,geoff boycott got excited and said that yuvraj has scored 210 runs in an over...,wat base system is he using??????
36 becomes 210
No. of digits has increased, so definitely the base system is
Besides, there is only 1 zero at the end of it. So the base system is a factor of 36 (2,3,4,6 -> less than 10), and cannot be 3 or 6 (both would leave 2 zeroes). It cant be 2 either coz base system of 2 consists of only 1s and 0s.
We are left with 4. So geoff boycotts base system is 4
How to derive total no of Squares in a chess board?
myalterego Sayshow to derive total no of squares in a chess board?
1^2+2^2+3^2+______+8^2=204
deepak_pgi Says1^2+2^2+3^2+______+8^2=204
Thanks. But I meant to ask why the approach works ..?
myalterego SaysThanks. But I meant to ask why the approach works ..?
Try to visualise yourself.
There is just one big square of dimension 8 x 8. -> 1^2
There are four squares of dimension 7 x 7 -> 2^2
.
..
...
.....
If you consider each small square of dimension 1 x 1 , there are 8 squares in each row, and 8 in each column. So, 8^2.
Add all of them ... 😃
myalterego SaysThanks. But I meant to ask why the approach works ..?
There will be squares with side as 1,2,3,______and 8.
Now the number of squares with side as 8= 1^2
the number of squares with side as 7= 2^2
the number of squares with side as 6=3^2__________
______________________________________________
the number of squares with side as 1=8^2
Hence the total number of squares will be
1^2+2^2+3^2+______+8^2=204
myalterego SaysThanks. But I meant to ask why the approach works ..?
Try to visualise yourself.
There is just one big square of dimension 8 x 8. -> 1^2
There are four squares of dimension 4 x 4 ->2^2
.
..
...
.....
If you consider each small square of dimension 1 x 1 , there are 8 squares in each row, and 8 in each column. So, 8^2.
Add all of them ... :)
Mistake in the bold line.....
Approach is this.....
8x8 squares---> 1^2
7x7 squares---> 2^2
6x6 squares---> 3^2
.
.
.
1x1 squares---> 8^2
How many integre values of x and y satisfy the expression of 4x+7y=3 where xa.284 b.285 c.286 .d.None of these .
The ans given is 285.Can someone explain the solution?
How many integre values of x and y satisfy the expression of 4x+7y=3 where xa.284 b.285 c.286 .d.None of these .
The ans given is 285.Can someone explain the solution?
4x+7y=3
x=(3-7y)/4
For integer values of x, we'll have to put
y=1,5,9, and so on, or y= -3,-7,-11...and so on.
case 1 )
For y=1,5,9... x=-1,-8,-15...
So, x will reach the value 1000 first.
Consider 1+7m
case 2 )
For y=-3, -7, -11... x=6,13,20...
So, again x will reach the value 1000 first.
Consider, 6+7n
Thus, total integer values = 143+142=285