5 Marks Questions

Question 15 Marks

If from a two-digit number, we subtract the number formed by reversing its digits then the result so obtained is a perfect cube. How many such numbers are possible? Write all of them.

Answer

Let ab be any two-digit number.
Then, the digit formed by reversing it digits is ba.
Now, $ab - ba = (10a + b) - (10b + a)$
$= (10a - a) + (b - 10b)$
$= 9a - 9b = 9(a - b)$
Further, since ab - ba is a perfect cube and is a multiple of $9$.
The possible value of $a - b$ is $3$. i.e. $a = b + 3$
Here, b can take value from $0$ to $6$.
Hence, possible numbers are as follow.
For $b = 0, a = 3$, i.e. $30$
For $b = 1, a = 4$, i.e. $41$
For $b = 2, a = 5$, i.e. $52$
For $b = 3, a = 6$, i.e. $63$
For $b = 4, a = 7$, i.e. $74$
For $b = 5, a = 8$, i.e. $85$
For $b = 6, a = 9$, i.e. $96$

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Question 25 Marks

Work out the following multiplication.
$\underline{\ \ \ \ \ 12345679\\\ \ \ \ \ \ \ \ \ \times\ \ \ \ \ 9 }\\\underline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }$
Use the result to answer the following questions.
$a.$ What will be $12345679 \times 45$?
$b.$ What will be $12345679 \times 63$?
$c.$ By what number should $12345679$ be multiplied to get $888888888$?
$d.$ By what number should $12345679$ be multiplied to get $999999999$?

Answer

which is actually the unit digit of the product $9 \times 9$.
Also, total no of digits in the multiplier is $90^\circ $
$a.$ Here, multiplier is $45$ whose sum of digits is $9$.
Thus, by conjecture we conclude that the product consists of digits $5$ only as unit’s digit of $9$ $\times 5$ is $5$.
$123456789$ $\times 45 = 555555555$
$b.$ Here, sum of the digits of the multiplier is $9$ and unit’s digit of the product is $9$ and unit’s digit of the products of $3 \times 9$ is $7$.
$12345679 \times 63 = 777777777$
$c.$ We have to obtain the number $8$ in the product, we should multiply the given number $123456789$ by $72$ as sum of its digits is $9$ and $2 \times 9$ has only digit as $8$.
$d.$ To get the number $9$ in the product we have to find out a two$-$digit number whose sum of digits is $9$ and the product of its unit’s digit with the given number is $9$, such number is $81$.

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Question 35 Marks

$\underline{\\ \ \ \ \text{8}\ \ \text{A} \ \ \text{B}\ \ \text{C}\\\text{-}\ \ \text{A}\ \ \text{B} \ \ \text{C}\ \ \ \text{5}}\\\underline{\ \ \ \ \text{D}\ \ \ 4\ \ \ 8\ \ \ 8}$

Answer

In the ones column, $C - 5 = 8$
Obviously, $13 - 5 = 8$ so, $C = 3$
In he ten’s column $B - (C + 1) = 8$
$B = 8 + C + 1$
$B = 8 + 3 + 1$
$B = 12$ i.e., $B = 2$
In the hundred’s column, $A − (B + 1) = 4$
$A = 4 + B + 1$
$A = 4 + 2 + 1 = 7$
In the thousand’s column, $8 - A = ?$
$8 − 7 = D$
$D = 1$
Hence, $A = 7, B = 2, C = 3$ and $D = 1$

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Question 45 Marks

Find the least value that must be given to number a so that the number $91876a2$ is divisible by $8$.

Answer

Given, $91876a2$ is divisible by $8$.
Since, we know that, if a number is divisible by $8$, then the number formed by last $3$ digits is divisible by $8$.
So, $6a2$ is divisible by $8$. Here, a can take values from $0$ to $9$.
For $a = 0, 602$ is not divisible by $8$. For $a = 1, 612$, which is not divisible by $8$.
For $a = 3, 632$ is divisible by $8$.
Hence, the minimum value of a is $3$ to make $91876a2$ divisible by $8$

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