If $a$ is the digit in unit place of the sum $1! + 2! + \dots + 39!, b$ is the digit in unit place of $(16)^{256}$ and $c$ is the digit in unit place of $(3)^{400},$ then $a + b + c$ is equal to

11

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

9

No worries! We‘ve got your back. Try BYJU‘S free classes today!

12

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $10$
We know, for $n \geq 5,$
$n!$ ends with $0.$
Now, $1! + 2! + 3! + 4! = 33$
$∴ a = 3$

$(16)^{n}$ always ends with $6, \forall n \in N$
So, $b = 6$

$3^{400} = (81)^{100} = (1 + 80)^{100}$
$= 1 +^{100} C_{1} \cdot 80 +^{100} C_{2} \cdot 80^{2} + \dots +^{100} C_{100} \cdot 80^{100}$
$= 1 + 100 k$ for some $k \in N$
$∴ c = 1$
Hence, $a + b + c = 3 + 6 + 1 = 10$

Suggest Corrections

Similar questions

Sum of digit of two number is 8. The digit in tens place is thrice the digit in unit place in unit place. Find the numbers?

10

The sum of the digits of a two-digit number is $9.$ If the digit in tens place is twice the digit in the units place, then the number is:

Sum of digits of two digits number is 8. The digit in tens place is thrice the digit in unit place. Find the digit.

Join BYJU'S Learning Program