Question

$(3132!{)}_{10}=(x{)}_{34}$, then what will be the number of consecutive zeros at the end of 'x'?

• A. 124
• B. 167
• C. 194
• D. None of these

Answer 1

The correct option is C 194

In base 34, 10 means 34. In base 10, 10 is obtained by multiplying 2 and 5. In base 34, it is obtained by multiplying 2 and 17. The number of consecutive zeroes in base 34 at the end of the number is same as the number of 2's and 17's in 3132!. Since the number of 2's is much more than the number of 17's, so we count the number of 17's in 3132!
Maximum power of 17 in $3132!=\left[\frac{3132}{17}\right]+\left[\frac{3132}{{17}^2}\right]$
= 184 + 10 = 194.