If z is the product of integers from 1 to 51, inclusive, what is the

Question

If z is the product of integers from 1 to 51, inclusive, what is the greatest value of k + m, where k and m are integers greater than 1, for which  45^k*32^m  is a factor of z?

A. 9
B. 20
C. 21
D. 23
E. 59

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GMATinsight · Answer

45^k*32^m = 2^{5m}*3^{2k}*5^k  is factors of 51!

51! = 2^{47}*3^{23}*5^{12}

i.e.  k_{max} = 11  becuase 2k≤23
i.e.  m_{max} = 9  becuase 5m≤47

i.e.  (k+m)_{max} = 11+9 = 20

Answer: Option B

raevant · Answer

Calculate the power of prime factors in a factorial:
Divide the highest number with increasing powers of the prime factor and add the quotients
eg. 51/2^1 + 51/2^2 + 51/2^3 + 51/2^4 + 51/2^5
= 25+12+6+3+1 = 47
Hence highest power of 2 that is a factor of 51 is 2^47
Similarly,
3^23 and 5^12
Now 5m =< 47 so m=9
and 3^23 * 5^12 = (9^11 x 3) * 5^12
Taking the limiting power gives 45^11
Hence k + m = 11 + 9 = 20

yashikaaggarwal · Answer

51! = Z

45^k + 32^m is a factor of z
45 = 3^2*5^1
32 = 2^5

We just have to determine no. Of 2s,3s and 5s value in 51!
=> 51/5 = 10/5 = 2
No. of 5 = 10+2 = 12

=> 51/3 = 17/3 = 5/3
No. of 3 = 17+5+1 = 23

=> 51/2 = 25/2 = 12/2 = 6/2 = 3/2 
No. of 2 = 25+12+6+3+1 = 47

M has to be a multiple of 5
The closest multiple of 5 to 47 = 45 = 5*9 = M = 9
32^m = 2^5m = 2^5*9 = 32^9 -----------(1)
45^k = 5^k and 3^2k

3^2K has to be a multiple of 2
The closest multiple of 2 to 23 = 22 = 2*11 = K = 11
3^2k = 3^2*11 = 3^22 -----------(2)

5^K has to be similar to that of 3^2k value
5^11 -----------(3)

Therefore K = 11 and M = 9
K+M = 11+9 = 20

Answer is B

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CesarAMSTech · Answer

45^k.32^m = 3^{2k}.5^k.2^{5m} so im going to count the number of factors of 2, 3 and 5 contained on 51! #of factors of 2 \frac{51}{2} = 25 , \frac{51}{2^2} = 12 , \frac{51}{2^3} = 6 , \frac{51}{2^4} = 3 , \frac{51}{2^5} = 1 Total = 25+12+6+3+1 = 47 #of factors of 3 \frac{51}{3} = 17 , \frac{51}{3^2} = 5 , \frac{51}{3^3} = 1 Total = 17+5+1 = 23 #of factors of 5 \frac{51}{5} = 10 , \frac{51}{5^2} = 2 Total = 10+2 = 12 So 51! = 2^{47}.3^{23}.5^{12}.(Integer) and we can compare term by term with 2^{5m}.3^{2k}.5^k Then 5m <= 47 , m(max) = 9 --> 2k <= 23 , k<=11.5 or --> k<= 12 k(max) = 11 Therefore, (m+k)max = 9+11 = 20

Superman249 · Answer

IMO:- B

45^k * 32 ^m = 3^2k * 5^k * 2^5m

Max Power of 2 in 51! = Greatest integer function (51/2) + Greatest integer function (51/4) + Greatest integer function (51/8)+ Greatest integer function (51/16) + Greatest integer function (51/32) = 25 + 12 + 6 + 3 +1 = 47

Max Power of 3 in 51! = Greatest integer function (51/3) + Greatest integer function (51/9) + Greatest integer function (51/27) = 17 + 5 + 1 = 23

Max Power of 5 in 51! = Greatest integer function (51/5) + Greatest integer function (51/25) = 10 + 2 =12

equating with 3^2k * 5^k * 2^ 5m => 2k multiple must be even no.; while 23 is not even; Hence the max value of 2k can be 22; So max k will be 11

While 5m must be multiple of 5; while 45 can be max; so max m will be 9

k+m (max)= 20

bumpbot · Answer

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If z is the product of integers from 1 to 51, inclusive, what is the

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