How many consecutive zeros are there at the end of the integer obtained in the product 1^2 \times 2^4 \times 3^6 \times 4^8 \times \dots \times 25^{50}?
- (a) 50
- (b) 55
- (c) 100
- (d) 200 ✓ UPSC's answer
Why the answer is (d)
• The product is $P = \prod_{k=1}^{25} k^{2k}$. The number of trailing zeros is determined by the number of pairs of prime factors 2 and 5, which is $\min(v_2(P), v_5(P))$. Since factors of 2 are more abundant than factors of 5, we calculate $v_5(P)$.
• The exponent of 5 in $P$ is the sum of the exponents of 5 in each term $k^{2k}$: $v_5(P) = \sum_{k=1}^{25} 2k \cdot v_5(k)$.
• Only terms where $k$ is a multiple of 5 contribute to this sum. The relevant values for $k$ are 5, 10, 15, 20, and 25.
• For $k=5, 10, 15, 20$, $v_5(k)=1$. The contribution is $2(5)(1) + 2(10)(1) + 2(15)(1) + 2(20)(1) = 10 + 20 + 30 + 40 = 100$.
• For $k=25$, $v_5(25)=2$ (since $25=5^2$). The contribution is $2(25)(2) = 100$.
• The total number of factors of 5 is $100 + 100 = 200$. Since $v_2(P) > 200$, the number of trailing zeros is 200.
Why the other options are wrong
- (a) 50
- Option 50 is incorrect because it significantly underestimates the total count of factor 5s contributed by the higher powers and larger bases in the product.
- (b) 55
- Option 55 is incorrect because it fails to account for the cumulative contribution of factor 5s from all multiples of 5 up to 25, particularly the double contribution from 25.
- (c) 100
- Option 100 is incorrect because it likely accounts for only the single factor of 5 in each multiple of 5 (5, 10, 15, 20, 25) without correctly weighting the exponent $2k$ or the double factor in 25.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2024, held on 16 June 2024. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.