@gs4890 said:It's satisfying the property of prime no's. Try it out by substituting any of them. Shouldn't the number be 5491 ?
Dude for eg: I consider 17. 17 can be LCM of (1,17). So its not satisfying the condition. Where I'm missing?
@gs4890 said:It's satisfying the property of prime no's. Try it out by substituting any of them. Shouldn't the number be 5491 ?
the question is asking the sum of 2 two digit nos.
@Subhashdec2 said:abcd-ab+cd=5481ab*100+cd-ab+cd=548199ab+2cd=5481cd has to be a multiple of 9possible sol-54,1872
answer is 73
@sunnychopra89 said:Dude for eg: I consider 17. 17 can be LCM of (1,17). So its not satisfying the condition. Where I'm missing?
Yes, it is satisfying. There's just one way if the no's has to be distinct
find the least value of n such that n! has exactly 2394 zero's .
@gs4890 said:Yes, it is satisfying. There's just one way if the no's has to be distinct
Bhai kya ho raha hai mujhe.....I'm not able 2 understand what u r trying to say....Plz could u show me with an eg......coz according to me non prime numbers shuld be the answer. eg; 24 can be LCM of (2,12) (3,8) (2,8,12). So total ways 3 (which is > 2) hence satisfies. But 17 LCM of (1,17) how is it satisfying?
In a 50 digit +ve no., all the digits are 4 except for the nth digit. This no. is divisible by 13 for some choice of that nth digit. How many possible values can n have?
Mr Roy found an old bill which shows that he brought 10 hens for Rs_67.92_ the first and the last digits were missing. what was the cost of each hen
??
??
@ravi.theja said:find the least value of n such that n! has exactly 2394 zero's .
2394*4=9576
9576 has 1915+383+76+15+3=2392
9580 has 1916+383+76+15+3=2393
9585 has 1917+383+76+15+3=2394
9585??
@sunnychopra89 said:In a 50 digit +ve no., all the digits are 4 except for the nth digit. This no. is divisible by 13 for some choice of that nth digit. How many possible values can n have?
is it 1??
case 1)
444-44n mod 13
4-n mod 13
n=4
case 2
444-4n4 mod 13
2-10-10n-4 mod 13
1-10n mod 13
1+3n mod 13
n=4
case 3)
444-n44 mod 13
2-9n-5 mod 13
4n+10 mod 13
n=4
case 4)
44n-444
n-4 mod 13
n=4
case5)
4n4-444 mod 13
36+10n+4-2 mod 13
38+10n mod 13
12+10n mod 13
n=4
case 6)
n44-444
100(n-4) mod 13
9n-36 mod 13
9n+3
n=4
if it comes at 49th position or 50th position
n4 should be divisible by 13
not possible
4n should be divisble by 13
no number is possible
so i guess only n=4 is possible
444-44n mod 13
4-n mod 13
n=4
case 2
444-4n4 mod 13
2-10-10n-4 mod 13
1-10n mod 13
1+3n mod 13
n=4
case 3)
444-n44 mod 13
2-9n-5 mod 13
4n+10 mod 13
n=4
case 4)
44n-444
n-4 mod 13
n=4
case5)
4n4-444 mod 13
36+10n+4-2 mod 13
38+10n mod 13
12+10n mod 13
n=4
case 6)
n44-444
100(n-4) mod 13
9n-36 mod 13
9n+3
n=4
if it comes at 49th position or 50th position
n4 should be divisible by 13
not possible
4n should be divisble by 13
no number is possible
so i guess only n=4 is possible
@Subhashdec2 said:is it 1??case 1)444-44n mod 134-n mod 13n=4case 2444-4n4 mod 132-10-10n-4 mod 131-10n mod 131+3n mod 13n=4case 3)444-n44 mod 132-9n-5 mod 134n+10 mod 13n=4case 4)44n-444n-4 mod 13n=4case5)4n4-444 mod 1336+10n+4-2 mod 1338+10n mod 1312+10n mod 13n=4case 6)n44-444100(n-4) mod 139n-36 mod 139n+3n=4if it comes at 49th position or 50th positionn4 should be divisible by 13 not possible4n should be divisble by 13no number is possibleso i guess only n=4 is possible
OA is 24.....:(
please tell me how can I find the quotient and remainder x^4-x^3+x^2-1 is divided by x^5-1 using factor theorem
find the value of 1-2-3+2-3-4+..+ upto 100 terms.
@sunnychopra89 said:Bhai kya ho raha hai mujhe.....I'm not able 2 understand what u r trying to say....Plz could u show me with an eg......coz according to me non prime numbers shuld be the answer. eg; 24 can be LCM of (2,12) (3,8) (2,8,12). So total ways 3 (which is > 2) hence satisfies. But 17 LCM of (1,17) how is it satisfying?
question aachi tarah pado. 1 he possible condition hain (1,17) & that is exactly what the Q is saying.
@ravi.theja said:find the least value of n such that n! has exactly 2394 zero's .
best approach will be to solve using options
@vbhvgupta said:if a times the ath term of an ap is equal to the b times the bth term, find (a+b)th term?1. 0 3. a^2 + b^22. a^2 - b^2 4. a - b5. 1
0 ?
@krittikau said:please tell me how can I find the quotient and remainder x^4-x^3+x^2-1 is divided by x^5-1 using factor theorem
the way i wld do it is put a random value of x, say 2 and then find the remainder and quotient and after that substitute 2 in the options and the one which satisfies wld be my answer....
@gs4890 said:question aachi tarah pado. 1 he possible condition hain (1,17) & that is exactly what the Q is saying. best approach will be to solve using options 0 ?
bhai..any other method?/ other than gng frm options??
@vbhvgupta said:find the value of 1-2-3+2-3-4+..+ upto 100 terms.
sum of (-4,-5,........-36) + 34
=> -626 ?
=> -626 ?
What is the no. of +ve integer triplets(a,b,c) that satisfy abc=4(a+b+c) with a