The digit at the tens place in the following expression is $(1!) + (2!)^{2} + (3!)^{3} + (4!)^{4} + (5!)^{5} + \dots (111!)^{111}$

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Solution

The correct option is C 9
The last two digits of (1!) = 01
The last two digits of $(2!)^{2}$ = 04
The last two digits of $(3!)^{3}$ = 16
The last two digits of $(4!)^{4}$ = 76
The last two digits of $(5!)^{5}$ = 00
Now, we can conclude that the last two digits of the higher numbers e.g., $(6!)^{6}, (7!)^{7}, (8!)^{8}$ . . . . etc. are ''00''. So the last two digits of the whole expression = 9
(Since 01 + 04 + 16 + 76 + 00 = 97)

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Similar questions

Q. What will be the tens place digit in the following expression:
1! + 2! + 3! + ... + 70!?

Question 3 (iii)
Simplify:

$\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}$

Consider a binary operation $*$ on set{1,2,3,4,5} given by the following multiplication table:
(i)Compute $(2 * 3) * 4$ and $2 * (3 * 4)$

$\begin{matrix} * & 1 & 2 & 3 & 4 & 5 1 & 1 & 1 & 1 & 1 & 1 2 & 1 & 2 & 1 & 2 & 1 3 & 1 & 1 & 3 & 1 & 1 4 & 1 & 2 & 1 & 4 & 1 5 & 1 & 1 & 1 & 1 & 5 \end{matrix}$

(ii) Is $*$ commutative?

$\begin{matrix} * & 1 & 2 & 3 & 4 & 5 1 & 1 & 1 & 1 & 1 & 1 2 & 1 & 2 & 1 & 2 & 1 3 & 1 & 1 & 3 & 1 & 1 4 & 1 & 2 & 1 & 4 & 1 5 & 1 & 1 & 1 & 1 & 5 \end{matrix}$

(iii)Compute $(2 * 3) \times (4 a s t 5)$
$\begin{matrix} * & 1 & 2 & 3 & 4 & 5 1 & 1 & 1 & 1 & 1 & 1 2 & 1 & 2 & 1 & 2 & 1 3 & 1 & 1 & 3 & 1 & 1 4 & 1 & 2 & 1 & 4 & 1 5 & 1 & 1 & 1 & 1 & 5 \end{matrix}$

Q. Observe the following pattern
2² − 1² = 2 + 1
3² − 2² = 3 + 2
4² − 3² = 4 + 3
5² − 4² = 5 + 4
and find the value of
(i) 100² − 99²
(ii) 111² − 109²
(iii) 99² − 96²

Question 3 (iii)
Simplify:

$\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}$

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The digit at the tens place in the following expression is (1!)+(2!)2+(3!)3+(4!)4+(5!)5+⋯(111!)111

The digit at the tens place in the following expression is $(1!) + (2!)^{2} + (3!)^{3} + (4!)^{4} + (5!)^{5} + \dots (111!)^{111}$