Latest CAT Aptitude Question SOLUTION: If 50 × 51 × 52 × 53 ×…× 200 is completely divisible by 9^a, then find the maximum possible value of ‘a’. Its answer is 37. But I don't know how ??

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If 50 × 51 × 52 × 53 ×…× 200 is completely divisible by 9^a, then find the maximum possible value of ‘a’. Its answer is 37. But I don't know how ??

Asked In CAT Sachin M (10 years ago)

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Saraswathy  (10 years ago)

50 × 51 × 52 × 53 ×…× 200 => 200! - 49!

9 => 3^2
[9 is not a prime factor... So we have to find the highest power of 3... From that we can calculate power of 9...]
No. of 3's in 200! is
200/3^1 + 200/3^2 + 200/3^3 + 200^3^4 = 66+22+7+2 = 97
No. of 3's in 49! is
49/3^1 + 49/3^2 + 49/3^3 = 16+5+1 = 22
No. of 3's in 50 × 51 × 52 × 53 ×…× 200 is (97-22) = 75
We have to find the highest power of 9 not 3 ... So 75/2 = 37

Ans : 37

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RAKESH  (10 years ago)

50 × 51 × 52 × 53 ×…× 200 = 200! / 49!

9^a = 3^2a
no. of 3's in 200! =
[200/3^1] + [200/3^2] + [200/3^3] + [200^3^4] = 66+22+7+2 = 97
no. of 3's in 49! =
[49/3^1] + [49/3^2] + [49/3^3] = 16+5+1 = 22

no. of 3's in 50 × 51 × 52 × 53 ×…× 200 = 200! / 49! = 3^97/3^22 = 3^75
for this to be completely divisible by 9^a = 3^2a , value of 2a should less or equal to 75
=> 2a = 75
=> a = 37

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