jlui4477 wrote:

The largest number amongst the following that will perfectly divide (101^100) - 1 is

a) 100

b) 10,000

c) 100^100

d) 100,000

Confirm that it's B!

The problem can be understood this way: find the largest divisor of 101^100 -1 among the following.

let's express 101^100 this way:

101^100= (100+1)^100 = 100^100*1^0+ A*100^99*1^1+ B*100^98*1^2 + ........+ Z*100^98*1^2 + Y*100^1*1^99+ T*100^0*1^100

The rule of expression is : the power of 100 reduces from 100 to 0 WHEREAS the power of 1 increases from 0 to 100

A, B, Z..... are coefficients which obeys some strict rule of the so-called Pascal triangle. From the Pascal triangle rule, Y (the last coefficient) is 100 and T=1

The remainder of this expression depends on the

2nd last term, (since the last term T*100^0*1^100 =1 . This last term is already terminated by -1 )

the second last term is Y*100^1*1^99= Y*100^1 ...Y=100 ---> Y*100= 10000.

Even the smallest term of the expression is perfectly divided by 10000 ---> the whole expression is the same.

The explanation looks lengthy but the logical processing is prompt if we know the generic expression of (a+b)^n