JEE Advanced 2016 (8)

Number Theory Level 4

For how many positive integers $x$ less than $4032$ is $x^{2} - 20$ divisible by 16 and $x^{2} - 16$ divisible by 20?

Source: OMO Spring 2016

405 400 402 403

1 solution

Aryan Goyat
Apr 7, 2016

x^(2)-20=16m and x^(2)-16=20n

-->x=2y

where y---0 to 2016

replacing x by y

y^(2)=4m+4+1

y^(2)=4n+4+n

now clearly visible n is of form 4x+1

y^(2)=20x+9

now this implies unit digit of y^(2) is 9 or of y is 7,3.

now we also find that any no of this unit digit is of form 20x+9

ie we just have to calculate total numbers with unit digit 7,3 in no from 0-2016.

Can you please explain why you did x = 2y ? Also can you tell how n is of form 4x + 1 ?

Priyadarshini R - 5 years, 2 months ago

from my first two eqn x^(2)-20=16m and x^(2)-16=20n it is clear that x^2 is a multiple of 4

ie x is a multiple of 2 so we can write it in form x=2y

y^(2)=4m+4+1

y^(2)=4n+4+n

equating the above we get

4m+4+1=4n+4+n n=4(m-n) +1=4x+1

aryan goyat - 5 years, 2 months ago