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Q13
(CISF/2020)
Miscellaneous & General Knowledge › Important Days, Places & Events
Consider the following number :
35 x 55 x 610 x 106 x 1512 x 1215 x 2573
What is the number of consecutive zeros at the end of the number given above ?
Result
Your answer:
—
·
Correct:
B
Explanation
To determine the number of consecutive zeros at the end of a product, we must find the number of pairs of factors 2 and 5, as 2 × 5 = 10. The number of trailing zeros is equal to the minimum of the total exponents of 2 and 5 in the prime factorization of the expression.
Breaking down the given number into prime factors:
- 35: (No factors of 2 or 5)
- 55: 55 (Count of 5s: 5)
- 610: (2 × 3)10 = 210 × 310 (Count of 2s: 10)
- 106: (2 × 5)6 = 26 × 56 (Count of 2s: 6, Count of 5s: 6)
- 1512: (3 × 5)12 = 312 × 512 (Count of 5s: 12)
- 1215: (22 × 3)15 = 230 × 315 (Count of 2s: 30)
- 2573: (52)73 = 5146 (Count of 5s: 146)
Total count of 2s = 10 + 6 + 30 = 46
Total count of 5s = 5 + 6 + 12 + 146 = 169
The number of trailing zeros is min(46, 169) = 46.
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