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If a, b, c, d, e, and f represent different prime numbers each, which of the following may represent the value of 16! ?
A. a^15 x b^6 x c^3 x d^2 x e^2 x f
B. a^15 x b^5 x c^3 x d^2 x e x f^2
C. a^15 x b^6 x c^3 x d^2 x e x f
D. a^8 x b^5 x c^3 x d^2 x e x f
E. a^8 x b^5 x c^2 x d^2 x e x f
A. a^15 x b^6 x c^3 x d^2 x e^2 x f
B. a^15 x b^5 x c^3 x d^2 x e x f^2
C. a^15 x b^6 x c^3 x d^2 x e x f
D. a^8 x b^5 x c^3 x d^2 x e x f
E. a^8 x b^5 x c^2 x d^2 x e x f
28 Jan 2026, 23:32
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ExpertsGlobal5
16! will have the following primes: 2, 3, 5, 7, 11, 13.
The exponents of 11 and 13 are clearly 1 each, i.e. 2 primes will have exponent 1 and the exponent of 7 will be 2 (since 7 and 14 are present), i.e. one prime will have exponent 2. Thus, A and B are eliminated.
Let us look at the exponent of 2 (since 2 will have the highest exponent since it is the smallest prime)
For 2, the exponent is 16/2 (= 8) + 8/2 (= 4) + 4/2 (= 2) + 2/2 (= 1) = 8 + 4 + 2 + 1 = 15 => We will have a prime with exponent 15. Thus, D and E are eliminated.
Answer C
29 Jan 2026, 00:26
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16!= 16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1
2^4 * (3*5) * (13) *(2*7)*(13)*(2^2*3)*(11)*(2*5)*(3^2)*(2^3)*7*(2*3)*5*(2^2)*(3)*(2)
2^15* 3^6* 5^3 * 7^2* 11 * 13
now if we compare this with the given option then only choice C matches.
2^4 * (3*5) * (13) *(2*7)*(13)*(2^2*3)*(11)*(2*5)*(3^2)*(2^3)*7*(2*3)*5*(2^2)*(3)*(2)
2^15* 3^6* 5^3 * 7^2* 11 * 13
now if we compare this with the given option then only choice C matches.
ExpertsGlobal5
