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

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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\(45^k.32^m = 3^{2k}.5^k.2^{5m}\) 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, \((m+k)max = 9+11 = 20\)

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

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