1. For a positive integer n, let \({P_n}\) denote the product of the
digits of $n$ and \({S_n}\) denote the sum of the digits of $n$. The number of
integers between 10 and 1000 for which \({P_n} + {S_n} = n\)
a. 9
b. 100
c. 128
d. 145
Answer: A
Explanation:
If the number is a two digit number be 'ab', then
Given, $a × b + a + b = 10a + b$
$a × b = 9a$
$b = 9$
Therefore, for $b = 9$, the above condition satisfies. So all the two
digit numbers for which units digit is 9 are our solutions. They range
from 19 to 99. Total 9 numbers.
If the number is a thre digit number 'abc', then
$a × b × c + a + b + c= 100a + 10b + c$
$ \Rightarrow a \times b \times c = 99a + 9b$
$ \Rightarrow b \times c = 99 + 9 \times \dfrac{b}{c}$
We know that $b, c$ are single digit number and takes maximum of $9$ each.
So the product on the L.H.S is maximum $9 × 9 = 81$ but R.H.S is greater
than 99 + something.
$\therefore$ no three digit number exists.
$\therefore$ Total numbers = $9$
2. The digits of a three-digit number A are written in the reverse
order to form another three-digit number B. If B > A and B – A is perfectly
divisible by 7, then which of the following is necessarily true?
a. 100 < A < 299
b. 106 < A < 305
c. 112 < A < 311
d. 118 < A < 317
Answer: B
Explanation:
Let $A = abc$ the $B = cba$
Given, $B-A$ is exactly divisible by $7$.
Therefore, $100c + 10b + a -(100a + 10b + c)$ $= 99c- 99a = 99(c-a)$
So $99(c- a)$ should be divisible by $7$. We know that $99$ is not
divisible by $7$.
So $(c- a)$ should be divisible by $7$ which implies $c-a$ is $0, 7,
14...$
$c-a = 0$ is not possible as $c = a$ implies $A = B$ but given that $B
> A$
$c- a = 14$ is not possible as the maximum difference between $c$ and $a =
9- 1 = 8$ only.
So $c- a = 7$
If $c = 9$, $a = 2$
If $c = 8, a = 1$
$b$ can take any value from 0 to 9
Therefore, minimum value of $abc = 109$, maximum value $= 299$
From the given options, option B satisfies this.
3. M = abc is a three digit number and N = cba, if M > N and M - N +
396c = 990. Then how many values of M are more than 300.
a. 20
b. 30
c. 40
d. 200
Answer: B
Explanation:
From the given data,
$abc – cba + 396c = 990$
$100a + 10b + c – (100c + 10b + a) + 396c = 990$
$99a – 99c + 396c = 990$
Observe that each term is divisible by 99. So on dividing the above
expression by 99, we get
$a – c + 4c = 10$
$a + 3c = 10$
For $c = 1, a = 7$
$c = 2, a = 4$
$c = 3, a = 1$
'$b$' can take any value from 0 to 9
We have to find the value of M more than 300. So minimum value of '$a$'
should be 4.
So total possibilities are 402, 412, ...., 492 = 10 values
701, 711, ....., 791 = 10 values
So total values = $20$.
4. Consider four digit numbers for which the first two digits are equal
and the last two digits are also equal. How many such numbers are perfect
square?
a. 2
b. 4
c. 0
d. 1
Answer: D
Explanation:
Let the given number by $xxyy$
So $1000x + 100x + 10y + y = 1100x + 11y = 11(100x + y)$
So $11(100x + y)$ has to be perfect square.
But the given expression is a multiple of 11 so $100x + y$ should also be
a multiple of 11 and a multiple of a perfect square.
Therefore, $11(100x + y)$ $= 11 × (11 × k^2) = 121 × k^2$
Now give values for $k$ and check the first two digits and last two digits
are equal or not.
For $k = 1$, $11(100x + y) = 11 × (11 × 1^2) = 121$ which is a 3 digit number so
ruled out.
For $k = 2$, $11 × (11 × 2^2) = 484$ also ruled out.
. . . .
. . . .
For $k = 8$, $11 × (11 × 8^2) = 7744$
For $k = 9$, you get $9801$. Ruled out.
If you further increase the value of $k$, the given product becomes 5
digit number.
So only one value is possible. That is $7744$.
5. By which number the expression $\dfrac{200!}{12^{100}}$ should
be multiplied so that the given expression becomes an integer
a. 216
b. 200
c. 64
d. 9
Answer: A
Explanation:
$12^{100}$ = $4^{100}\times3^{100}$ = $2^{200}\times3^{100}$
Now we have to find the maximum power of 2 and 3 in numerator.
$\dfrac{200!}{12^{200}}=\dfrac{2^{197}\times3^{97}\times... .
}{2^{200}\times3^{100}}$
To divide the numerator, we need to multiply it with
$2^{3}\times3^{3}=216$
6. What is the maximum power of 3 in the expansion of 1! ×
2! × 3! × . . . . × 100!
a. 11
b. 13
c. 14
d. 18
Answer: B
Explanation:
Given 1! × 2! × 3! × . . . . × 100!. We rewrite this expression
by writing as 1100 × 299 ×
398 × . . . × 1001
This is possible as each term contains 1. From 2! on wards each term
contains 2. So they are total 99. Similarly, only last term contains 100.
So it has power of 1.
Now we have to calculate number of 3's.
398, 695, 992. . ., 992
Observer here the power and base sum = 101. So it is easy to form this
series.
Total powers of 3 = 98 + 95 + 92 + . . . . + 2
Number of terms = $\dfrac{{l - a}}{d} + 1$ = $\dfrac{{98 -
2}}{3} + 1$ = 33
Sum = $\dfrac{n}{2}\left( {a + l} \right)$
= $\dfrac{33}{2}\left( {98 + 2} \right)$ = 1650
Also, the powers of 9's, 18's, 27's contains another 3. They
contribute more number of 3's.
So 992, 1883, 2774. . ., 992
Number of terms = $\dfrac{{l - a}}{d} + 1$ = $\dfrac{{92 - 2}}{9} +
1$ = 11
Sum = $\dfrac{n}{2}\left( {a + l} \right)$
= $\dfrac{11}{2}\left( {2 + 92} \right)$ = 517
Also, the powers of 27's, 54's, 81's contains another 3.
So 2774, 5447, 8120
Their sum = 74 + 47 + 20 = 141
Finally, power of 81 contains another 3. So it contributes another 20.
Total = 1650 + 517 + 141 + 20 = 2328.
