Question

Write the quotient, when the sum of a 2-digit number 23 and number obtained by reversing the digits divided by
(i) 11
(ii) Sum of the digits.

Answer 1

Hint: To solve the question, at first we have to expand the two digit number 23 in the form\[10x+y\]. By obtaining the values of x and y we have to find out \[10y+x\] which is the number obtained by reversing the digits of the number 23. Then we will find out the sum 23 and the reversed number\[10y+x\]. Finally we must divide the obtained sum by 11 and then by the value of the sum of the digits of 23 that is the value of \[x+y\]to get the answers.

Complete step-by-step answer:
The given number is 23 and can be expanded in the form of \[10x+y\]as follows.
\[23=10\times 2+3\]
Here we get by comparing \[10x+y\]that \[x=3\]and\[y=2\]. Then the number obtained by reversing the digits of the number 23 is given by\[10y+x=10\times 3+2=32\].
The sum of 23 and the reversed number 32 is given by
\[23+32=55\]
When we divide 55 by 11, we know that 55 is product of factor of prime numbers 5 and 11 so, we can write 55 as $5\times 11$, that means\[\dfrac{55}{11}=\dfrac{5\times 11}{11}\]
On solving, we get 5 as quotient that is \[\dfrac{5\times 11}{11}=5\]
The sum of digits of 23 is given by
\[x+y=2+3=5\].
When we divide 55 by 5 we get 11 as quotient that means \[\dfrac{55}{5}=11\]
Therefore we got required answers 5 and 11 respectively.

Note: In alternative method we can interchange the ones and tens places of the number to get the number reversed that is if we have 54 as a number then interchange of the ones and tens place of number means we change the place value of 5 from tens to ones and place value of 4 from ones to tens, that is we have resulting number as 45. Try not to make any calculation mistakes.