Quiz 3

The four consecutive two-digit odd numbers will have (1, 3, 5, 7) or (3, 5, 7, 9) or

(5, 7, 9, 1) or (7, 9, 1, 3) as units digits. As the sum divided by 10 yields a perfect square, the sum is a multiple of 10. ∴ The units digits have to be (7, 9, 1, 3). Thus the four numbers will be (10x + 7), (10x + 9), (10x + 11) and (10x + 13), where 0 < x < 9 (as each of these numbers is a two digit number) Sum of these numbers = 40x + 40 = 40(x + 1) Now, 40(x + 1)/10 = 4(x + 1) is a perfect square As 4 is a perfect square (x + 1) is some perfect square < 10 x + 1 = 4, x = 3, and the four numbers are 37, 39, 41 and 43 x + 1 = 9, x = 8, and the four numbers are 87, 89, 91 and 93

Hence, option 3.

Let 10x + y be a two digit number, where x and y are positive single digit integers

and x > 0. Its reverse = 10y + x Now, 10y + x – 10x – y = 18

∴ 9(y – x) = 18 ∴ y – x = 2.

Thus y and x can be (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8) and (7, 9) ∴ Other than 13, there are 6 such numbers. Hence, option 2.

Consider options. As the number of employees is prime we can add the

numerator and denominator of ratios directly to find the number of employees. 1. Number of employees = 101 + 88 = 189

Number of employees = 189, which is not a prime number. ∴ Option 1 is eliminated. 2. Number of employees = 87 + 100 = 187

Number of employees = 187, which is not a prime number.

∴ Option 2 is eliminated. 3. Number of employees = 110 + 111 = 221

Number of employees = 221, which is not a prime number. ∴ Option 3 is eliminated. 4. Number of employees = 85 + 98 = 183

Number of employees = 183, which is not a prime number. ∴ Option 4 is eliminated. 5. Number of employees = 97 + 84 = 181

Number of employees = 181, which is a prime number.

∴ The ratio of employees = 97:84 Hence, option 5.

Formula: Remainder(A^n + B^n/ A+B )=0 When n=odd

Here Rem( $(15^{23}+23^{23})$ / 19)

N is odd

so , 15+23= 38

and 38 is divisible by 19

so Remainder: 0

We know that x = 16^3 + 17^3 + 18^3 + 19^3 = (16^3 + 19^3) + (17^3 + 18^3)

=>x=70k

=> Remainder, when divided by 70, is 0

p = (1 × 1!) + (2 × 2!) + (3 × 3!) + (4 × 4!) + ... + (10 × 10!)
Now, n × n! = [(n + 1) – 1] × n! = (n + 1)! – n!
∴ p = 2! – 1! + 3! – 2! + 4! – 3! + 5! – 4! +... + 11! – 10!
∴ p = 11! – 1! = 11! – 1
∴ p + 2 = 11! + 1
∴ p + 2 when divided by 11! leaves a remainder of 1.
Hence, option 3.

Let A = 100x + 10y + z (x ≠ 0, x, y, z are single digit numbers)
∴ B = 100z + 10y + x
∴ B – A = 99(z – x)
As (B – A) is divisible by 7 and 99 is not, (z – x) is divisible by 7
∴ z and x can have values (8, 1) or (9, 2)
y can have any value from 0 to 9.
A = 1y8 or 2y9
∴ Lowest possible value of A is 108 and the highest possible value of A is 299.
Hence, option 2.

Let O and E represent odd and even digits respectively.
∴ S can have digits of the form
O _ O _ E or O _ E _ O or E _ O _ O
Case 1: O _ O _ E
The first digit can be chosen in 3 ways out of 1, 3 and 5
The third can be chosen in 2 ways.
The fifth digit can be chosen in 2 ways after which the second and fourth digits
can be chosen in 2 ways.
∴ There are 3 × 2 × 2 × 3 = 24 ways in which this number can be written. 12 out
of these ways will have 2 in the rightmost position and 12 will have 4 in the
rightmost position.
∴ The sum of the rightmost digits in Case 1 = (12 × 2) + (12 × 4) = 72
Case 2: O _ E _ O
This number can again be written in 24 ways.
8 out of these ways will have 1 in the rightmost position, 8 will have 3 in the
rightmost position and 8 will have 5 in the rightmost position.
Thus the sum of the rightmost digits in Case 2 = (8 × 1) + (8 × 3) + (8 × 5) = 72
Case 3: E _ O _ O
This number can also be written in 24 ways.
As in Case 2, 8 out of these ways will have 1 in the rightmost position, 8 will have
3 in the rightmost position and 8 will have 5 in the rightmost position.
∴ The sum of the rightmost digits in Case 3 = (8 × 1) + (8 × 3) + (8 × 5) = 72
∴ The sum of the digits in the rightmost position of the numbers in S = 72 + 72 +
72 = 216
Hence, option 2.

302720 = 32720 × 102720
The rightmost non zero digit of 302720 will be the digit in the unit’s place of 32720.
3’s power cycle is 3, 9, 7, 1 and cyclicity is 4.
2720 = 680 × 4
∴ The digit in the unit’s place of 32720 is 1.
∴ The rightmost non-zero digit of 302720 is 1.
Hence, option 1.

n can be a 2 digit or a 3 digit number.
Case (I)
Let n be a 2 digit number.
Let n = 10x + y, where x and y are non-negative integers,
Pn = xy and Sn = x + y
Now, Pn + Sn = n
∴ xy + x + y = 10x + y
∴ xy = 9xy = 9
There are 9 two digit numbers (19, 29, 29, ... ,99) for which y = 9
Case (II)
Let n be a 3 digit number.
Let n = 100x + 10y + z, where x, y and z are non-negative integers,
Pn = xyz and Sn = x + y + z
Now, Pn + Sn = n
xyz + x + y + z = 100x + 10y + z
∴ xyz = 99x + 9y
∴ z = 99/y + 9/x
From the above expression, 0 < x, y < 9
But, we cannot find any value of x and y, for which z is a single digit number.
∴ There are no 3 digit numbers which satisfy Pn + Sn = n
Hence, option 4.