HCR's Series
Let there be two positive integers a & b such that
a!-b!=103×103!+102×102!+101×101!+ …………+3×3!+2×2!+1×1!
Find out the value of (a+b)
The answer is 105.
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2 solutions
1 0 3 ∗ 1 0 3 ! + 1 0 2 ∗ 1 0 2 ! … + 2 ∗ 2 ! + 1 ∗ 1 !
= 1 0 3 ∗ 1 0 3 ! + 1 0 2 ∗ 1 0 2 ! + … + 2 ∗ 2 ! + 1 ∗ 1 ! + 1 ! − 1 !
= 1 0 3 ∗ 1 0 3 ! + 1 0 2 ∗ 1 0 2 ! + … + 2 ∗ 2 ! + 2 ∗ 1 ! − 1 !
= 1 0 3 ∗ 1 0 3 ! + 1 0 2 ∗ 1 0 2 ! + … + 2 ∗ 2 ! + 2 ! − 1 !
= 1 0 3 ∗ 1 0 3 ! + 1 0 2 ∗ 1 0 2 ! + … + 3 ∗ 2 ! − 1 !
= 1 0 3 ∗ 1 0 3 ! + 1 0 2 ∗ 1 0 2 ! + … + 3 ! − 1 !
Continuing in this way,
= 1 0 3 ∗ 1 0 3 ! + 1 0 3 ! − 1 !
= 1 0 4 ∗ 1 0 3 ! − 1 !
= 1 0 4 ! − 1 !
Therefore, a = 1 0 4 , b = 1 and a + b = 1 0 5
103 * 103! = (104 - 1) * 103! = 104 * 103! - 103! = 104! - 103 !
that means,
102 * 102! = 103! - 102!
101 * 101! = 102! - 101!
................. .........
1 * 1! = 2! -1!
So, a! - b! = 103×103!+102×102!+101×101!+ …………+3×3!+2×2!+1×1! = 104! - 1!
Therefore a = 104 and b = 1
so, a + b =105