Identical twins in inverse order

Number Theory Level pending

Let a a and b b and 2 positive integers such that their product is 4032, and if we reverse the digits of a a , we obtain the number b b .

Given that the greatest common divisor of a a and b b is a factor of the difference a − b a-b , find the maximum possible value of between these 2 numbers, a a and b b .


The answer is 84.

Shaily Shivangi by Shaily Shivangi , 18, India

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1 solution

Shaily Shivangi
Feb 16, 2017

Here it is given, a×b= 4032 And, We know that 8!(i.e.1×2×3×4×5×6×7×8)=40320 .....(I) So if we remove 10(i.e.2×5), then 1×3×4×6×7×8=4032. .....(ii) Now if we form groups as (6×8) and (3×4×7) we would get: 84×48=4032. .....(iii) And the later part of the questions can be used to verify the answer as, HCF of a and b ( ie. 84 and 48) is 12. ....(IV) And difference between a and b - 84-48=36. ....(v), which has 12 as a factor.

And a is greater b as in the equation (v), b is subtracted from a and the difference is a positive number.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

The phrasing of the problem is convoluted, and it's not easy to understand what you're trying to express.

Can you clean up the problem so that others will find it easier to read?

Calvin Lin Staff - 4 years, 3 months ago

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