Question

Let != 1×2×3×.....×n. For example, 4!= 1×2×3×4 and 3! = 1×2×3.
Find the last digit of 9!+99!+999!+9999!+99999!.

Answer 1

Consider 9!

9! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9

In this product, 2 and 5 are multiplied too. The product of 2 and 5 is 10, thus even when the other numbers in 9! are multiplied with this 10 we get the last digit is 0, thus 9! ends in 0.

Consider 99!

In 99!, there are so many multiplies of 10 like 10, 20, 30, 40, 50, 60, 70, 80, 90. Thus 99! ends in many zeroes (more than 10 zeroes). Thus the last digit of 99! is also 0.

So do 999!, 9999! and 99999!.

The sum of two or more numbers ending in zero also ends in zero.

Thus 9! + 99! + 999! + 9999! + 99999! also ends in zero.

Hence 0 is the answer.

Answer 2

Heya!!!

9! + 99! + 999! + 9999! + 99999! =

9! { 1 + 11 + 111 + 1111 + 11111 }

=>

Multiple NR and DN by 9

9!/9 { 9 + 99 + 999 + 9999 + 99999 }

=>

9!/9 { (10 - 1 ) + ( 10² - 1 ) + ( 10³ - 1 ) +

( 10⁴ - 1 ) + ( 10^5 - 1 ) }

=>

9!/9 { 10 + 10² +10³ + 10⁴ + 10^5 - 5 }

=>

9!/9 { 11110 - 5 }

=>

9!/9 { 1111 05 }

=>

( 111105 ) × 8!

= 4479753600

So, it's last digit is zero ( 0 )