Let A denote the set of all 2-digit numbers i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the $2$-digit number be represented as $10a + b$,where $a \in \{1, 2, \dots, 9\}$ and $b \in \{0, 1, \dots, 9\}$.According to the problem,$10a

Vedclass · Accepted Answer

$44$

Vedclass · Answer

(C) Let the $2$-digit number be represented as $10a + b$,where $a \in \{1, 2, \dots, 9\}$ and $b \in \{0, 1, \dots, 9\}$.According to the problem,$10a + b = 4(a! + b!)$.Since $10a + b \leq 99$,we have $4(a! + b!) \leq 99$,which implies $a! + b! \leq 24.75$.This restricts $a$ and $b$ to values $\leq 4$ because $5! = 120 > 24$.Testing possible values for $a \in \{1, 2, 3, 4\}$ and $b \in \{0, 1, 2, 3, 4\}$:If $a=1$: $10(1) + b = 4(1! + b!)$ $\Rightarrow 10 + b = 4(1 + b!)$ $\Rightarrow 6 + b = 4b!$. For $b=2$,$6+2 = 4(2!) = 8$. So,$12$ is a solution.If $a=2$: $20 + b = 4(2! + b!)$ $\Rightarrow 20 + b = 8 + 4b!$ $\Rightarrow 12 + b = 4b!$. No integer solution for $b \in \{0, 1, 2, 3, 4\}$.If $a=3$: $30 + b = 4(3! + b!)$ $\Rightarrow 30 + b = 24 + 4b!$ $\Rightarrow 6 + b = 4b!$. For $b=2$,$6+2 = 4(2!) = 8$. So,$32$ is a solution.If $a=4$: $40 + b = 4(4! + b!) \Rightarrow 40 + b = 96 + 4b!$. No solution as $40+b The set $A = \{12, 32\}$.The sum of the numbers in $A$ is $12 + 32 = 44$.

Let $A$ denote the set of all $2$-digit numbers in base $10$ that are equal to four times the sum of the factorials of their digits. The sum of the numbers in $A$ is

Similar Questions

Let $S = \{(a, b) : a, b \in \mathbb{Z}, 0 \leq a, b \leq 18\}$. The number of elements $(x, y)$ in $S$ such that $3x + 4y + 5$ is divisible by $19$ is:

Let $X$ be the set consisting of the first $2018$ terms of the arithmetic progression $1, 6, 11, \dots$ and $Y$ be the set consisting of the first $2018$ terms of the arithmetic progression $9, 16, 23, \dots$. Then,the number of elements in the set $X \cup Y$ is:

In a class of $60$ students,$40$ opted for $NCC$,$30$ opted for $NSS$ and $20$ opted for both $NCC$ and $NSS$. If one of these students is selected at random,then the probability that the student selected has opted neither for $NCC$ nor for $NSS$ is

If $U$ is the universal set with $100$ elements; $A$ and $B$ are two sets such that $n(A)=50$,$n(B)=60$,and $n(A \cap B)=20$,then find $n(A^{\prime} \cap B^{\prime})$.

The number of $3$-digit numbers that are divisible by either $3$ or $4$ but not divisible by $48$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module