How do you prove that 5 to the 31st square minus 5 to the 29th square is divisible by 100?

Question

Format: Blank times 100

Answer 1

Proving that 100 is factor is sufficient.

(5^31)^2 - (5^29)^2 =

(5^31 + 5^29)(5^31 - 5^29) <--- difference of squares pattern

= 5^29 ( 5^2 + 1) * 5^29 ( 5^2 - 1) <-- factors out 5^29 out of each binomial

=5^29 (25 +1) * 5^29 * (25-1)

= 5^29 * 26 * 5^29 * 24

= (5*5^28) * (2 * 13) *( 5 * 5^28) *( 2 * 12) <--- 5^29 = 5*5^28 ; 26 = 2*13; 24 = 2*12

= 5*5^28 * 2 * 13 * 5 * 5^28 * 2 * 12 <-- per associative property, parenthesis dropped

= 5 * 2 * 5^28 * 13 * 5 * 2 * 5^28 * 12 <--- commutative property

= (5*2) * 5^28 * 13 * (5*2) * 5^28 * 12 <--- associative property

= 10 * 5^28 * 13 * 10 * 5^28 * 12

= 10 * 10 * 5^28 * 13 * 5^28 * 12 <-- commutative again

= (10 * 10) * 5^28 * 13 * 5^28 * 12 <--- associative again

= 100 * 5^28 * 13 * 5^28 * 12

therefore 100 is a factor , so the original expression is divisible by 100

the quotient shall be 5^56*156