Just another Trailing Zeroes Problem!

Number Theory Level 3

We are three (3) children in our family. The product of our ages is 7560, while the sum is 59. Let a be my oldest sibling's age and b for the second oldest.

Find the number of trailing zeroes in (a+b)!

You can also try this or My Other Problems .

The answer is 9.

2 solutions

Sunil Pradhan
Jul 4, 2014

abc = 7560 and a + b + c = 59

7560 = 9 × 7 × 5 × 4 × 3 × 2

by trial and error (7×3) + (5 × 4) + (9 × 2) = 21 + 20 + 18 = 59

a = 20, b = 20 then (a + b)! = 41!

to find number of zeroes at the end of product

41/5 = 8

8/5 = 1 (quotient less than 5)

Total zeros = 8 + 1 = 9

How do you get the no as 9,7,5,4,3,2

Nikhil Niki - 6 years, 8 months ago

there can be other solution for this problem also. a+b+c = 59 or 30 + 21 + 8 = 59,

so 30 +21 = 51.

51/5 = 10 10/5 = 2 (quotient less than 5)

Total zeros = 10+2 = 12

Jabid Ahamed - 6 years, 10 months ago

30, 21 and 8 cannot be their ages since the product of 30, 21 and 8 is not equal to 7560.

Jack Mamati - 6 years, 10 months ago

Saqib M
Aug 17, 2014

Well, you can see the sharer's age below the question itself ;), (beside the name), so one variable = 18 and if we check if whether 7560 is divisible by 18, then Viola!, it is. So, A+B = 41 and AB = 420.

Now we need to find trailing zeroes in 41!. For that we will have to find how many 5's come in 41!.

By Legendre's Theorem, $\left[ \frac { 41 }{ 5 } \right] +\left[ \frac { 41 }{ 25 } \right]$ = 8 +1 = 9

So wise :D

Astro Enthusiast - 6 years, 9 months ago

excellent visual power and too lucky

Viswakanth Kandala - 6 years, 8 months ago

nice one !

Henrison Sanchez - 6 years, 8 months ago

well reading your soln. made my day...:) gr8 one

Ritwik Jain - 5 years, 12 months ago

Just another Trailing Zeroes Problem!

The answer is 9.

2 solutions

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