Question 14 Marks
a, b, c are three different nonzero digits. Using these digits, different numbers are formed.
(1) If $a>b>c$ then the difference of three-digit number $a b c$ and the number obtained by reversing the digits is always divisible by
(a) 9 only$\quad$(b) 11 only$\quad$(c) both 9 and 11$\quad$(d) none of 9 and 11
(2) How many different three-digit numbers can be formed using the given digits it being given that repetition of digits is not allowed?
(a) 3$\quad$(b) 4$\quad$(c) 6$\quad$(d) 9
(3) The sum of all possible three-digit numbers formed above, is always divisible by
(a) 2 only$\quad$(b) 37 only$\quad$(c) 111 only$\quad$(d) each one of 2, 37 and 111
(4) Which of the following is not true?
(a) The number $a a b b$ is divisible by 11$\quad$
(b) The number $a b a b$ is divisible by 101$\quad$
(c) The number $a a b b c c$ is divisible by 11$\quad$
(d) The number $a b c a b c$ is divisible by 101
(1) If $a>b>c$ then the difference of three-digit number $a b c$ and the number obtained by reversing the digits is always divisible by
(a) 9 only$\quad$(b) 11 only$\quad$(c) both 9 and 11$\quad$(d) none of 9 and 11
(2) How many different three-digit numbers can be formed using the given digits it being given that repetition of digits is not allowed?
(a) 3$\quad$(b) 4$\quad$(c) 6$\quad$(d) 9
(3) The sum of all possible three-digit numbers formed above, is always divisible by
(a) 2 only$\quad$(b) 37 only$\quad$(c) 111 only$\quad$(d) each one of 2, 37 and 111
(4) Which of the following is not true?
(a) The number $a a b b$ is divisible by 11$\quad$
(b) The number $a b a b$ is divisible by 101$\quad$
(c) The number $a a b b c c$ is divisible by 11$\quad$
(d) The number $a b c a b c$ is divisible by 101
Answer
View full question & answer→(1) (C) both 9 and 11
Difference $=(100 a+10 b+c)-(100 c+10 b+a)=99 a-99 c=99(a-c)$, which is clearly divisible by 99 and hence by both 9 and 11.
(2) (C) 6
The different three-digit numbers that can be formed using the given digits are $a b c, a c b, b a c, b c a, c a b, c b a$, i.e., 6 in all.
(3) (D) each one of 2,37 and 111
$=222(a+b+c)$, which is clearly divisible by 2, 37 and 111 .
(4) (D) The number $a b c a b c$ is divisible by 101
$\begin{aligned} a a b b & \rightarrow 1000 a+100 a+10 b+b \\ & =1100 a+11 b=11(100 a+b), \text { which is clearly divisible by } 11 .\end{aligned}$
$\begin{aligned} a b a b & \rightarrow 1000 a+100 b+10 a+b \\ & =1010 a+101 b=101(10 a+b), \text { which is clearly divisible by } 101 .\end{aligned}$
$\begin{aligned} a a b b c c & \rightarrow 100000 a+10000 a+1000 b+100 b+10 c+c \\ & =110000 a+1100 b+11 c \\ & =11(10000 a+100 b+c), \text { which is clearly divisible by } 11 .\end{aligned}$
$\begin{aligned} a b c a b c & \rightarrow 100000 a+10000 b+1000 c+100 a+10 b+c \\ & =100100 a+10010 b+1001 c \\ & =1001(100 a+10 b+c), \text { which is clearly divisible by } 1001 \text { and not by } 101 .\end{aligned}$
Difference $=(100 a+10 b+c)-(100 c+10 b+a)=99 a-99 c=99(a-c)$, which is clearly divisible by 99 and hence by both 9 and 11.
(2) (C) 6
The different three-digit numbers that can be formed using the given digits are $a b c, a c b, b a c, b c a, c a b, c b a$, i.e., 6 in all.
(3) (D) each one of 2,37 and 111
$=222(a+b+c)$, which is clearly divisible by 2, 37 and 111 .
(4) (D) The number $a b c a b c$ is divisible by 101
$\begin{aligned} a a b b & \rightarrow 1000 a+100 a+10 b+b \\ & =1100 a+11 b=11(100 a+b), \text { which is clearly divisible by } 11 .\end{aligned}$
$\begin{aligned} a b a b & \rightarrow 1000 a+100 b+10 a+b \\ & =1010 a+101 b=101(10 a+b), \text { which is clearly divisible by } 101 .\end{aligned}$
$\begin{aligned} a a b b c c & \rightarrow 100000 a+10000 a+1000 b+100 b+10 c+c \\ & =110000 a+1100 b+11 c \\ & =11(10000 a+100 b+c), \text { which is clearly divisible by } 11 .\end{aligned}$
$\begin{aligned} a b c a b c & \rightarrow 100000 a+10000 b+1000 c+100 a+10 b+c \\ & =100100 a+10010 b+1001 c \\ & =1001(100 a+10 b+c), \text { which is clearly divisible by } 1001 \text { and not by } 101 .\end{aligned}$