Olympiad Mathematics - Playing with Numbers
Exam Duration: 45 Mins Total Questions : 20
The number (10n - 1) is divisible by 11 for
- (a)
n \(\epsilon \) N
- (b)
Odd values of n
- (c)
Even values of n
- (d)
n is the multiple of 11
(10n - 1) is divisible by 11 for even values of n as 102 - 1 = 99, 104 - 1 = 9999, 106 - 1 = 999999 etc, are divisible by 11
The values of A and 8 in the given addition respectively are __
2 3 A
+A 3 B
- (a)
4, 7
- (b)
7, 4
- (c)
5, 6
- (d)
6, 5
2 3 \(\boxed { 4 } \)
+\(\boxed { 4 } \) 3 \(\boxed { 7 } \)
________
6 \(\boxed { 7 } \) 1
________
The greatest value that must be given to x so that the number 7713 x 8 is divisible by 4 is __________
- (a)
1
- (b)
6
- (c)
8
- (d)
7
7713x8 is divisible by 4 if its last two digits is divisible by 4. So, we have x8 to be divisible by 4. Therefore, greatest value of x is 8.
If 1 A where A and 8 are single digit numbers, such that 8 - A = 3, then the values of A and 8 respectively are _________
x A
_____
B 6
______
- (a)
4, 5
- (b)
9, 6
- (c)
5, 4
- (d)
6, 9
We have,
1 \(\boxed { 6 } \) and 1 \(\boxed { 4 } \)
x \(\boxed { 6 } \) x \(\boxed { 4 } \)
____ ______
\(\boxed { 9 } \) 6 \(\boxed { 5 } \) 6
_____ ______
Since, B - A = 3
\(\therefore\)Possible values of A and Bare 6 and 9.
Suppose that the division N + 5 leaves a remainder of 4 and the division N + 2 leaves a remainder of 1. What must be the ones digit of N?
- (a)
7
- (b)
3
- (c)
9
- (d)
4
Given that the number 148101a095 is divisible by 11, where a IS single digit number, what are the possible values of a?
- (a)
4
- (b)
1
- (c)
7
- (d)
9
If 1481018095 is divisible by 11, then
(1 + 8 + 0 + 8 + 9) - (4 + 1 + 1 + 0 + 5) is either 0 or a multiple of 11
\(\therefore\)(a + 18) - 11 is either 0 or a multiple of 11
\(\therefore\)a + 7 = 0 or a multiple of 11
\(\therefore\)a + 7 = 11 \(\therefore\) a = 4
The largest natural number by which the product of three consecutive even natural numbers IS always divisible, is ______
- (a)
16
- (b)
24
- (c)
48
- (d)
96
The required number would be the product of three smallest even natural number i.e. 2 x 4 x 6 = 48.
If in a number, difference between the sum of digits at its odd places and that of digits at the even places is given 0, then the number is divisible by _______
- (a)
7
- (b)
9
- (c)
5
- (d)
11
A 5-digit number xy235 is divisible by 3 such that x + y < 5, where x and y are single digits, then possible values of (x, y) are ________
- (a)
(1,1)or(4,0)
- (b)
(1,1) or (2,0)
- (c)
(1,1) or (0,2)
- (d)
(2, 0) or (0, 2)
xy235 is divisible by 3 so sum of its digits is also divisible by 3.
\(\therefore\)x + y + 2 + 3 + 5 = x + y + 10 is divisible by 3 and x + y < 5.
So possible values of (x, y) are (1, 1) or (2,0).
If 1 A B and there is no carry on addition then the value of 8 is ________
C C A
______
6 9 7
______
- (a)
5
- (b)
4
- (c)
3
- (d)
2
1 \(\boxed { 4 } \) \(\boxed { 3 } \)
+ \(\boxed { 5 } \) \(\boxed { 5 } \) \(\boxed { 4 } \)
__________
6 9 7
__________
If N divided by 5 leaves a remainder of 3, then one's digit of N must be ________
- (a)
Either 3 or 6
- (b)
Either 3 or 8
- (c)
Either 8 or 1
- (d)
Either 8 or 6
N leaves remainder 3 when divided by 5.
\(\therefore\)(N - 3) is divisible by 5.
\(\therefore\)One's digit of N - 3 is either 0 or 5.
\(\therefore\)One's digit of N is either 3 or 8.
Given that the number 67y19 is divisible by 9, where y is a single digit, what is the least possible value of y?
- (a)
3
- (b)
9
- (c)
7
- (d)
4
67y19 is divisible by 9 so sum of its digits is also divisible by 9
6 + 7 + Y + 1 + 9 = 23 + y is divisible by 9.
So, least possible value of y = 4.
A 3-digit number 'eba' is divisible by 3 if ________
- (a)
a + 2b + e is divisible by 3
- (b)
2a + b + e is divisible by 3
- (c)
a + b + 2e is divisible by 3
- (d)
a + b + e is divisible by 3
If A B, then the value of B is _________
x A 3
_____
5 7 B
_____
- (a)
5
- (b)
2
- (c)
0
- (d)
4
\(\quad \boxed { 2 } \quad \boxed { 5 } \\ \times \boxed { 2 } \quad 3\)
__________
7 5
+ 5 0 0
__________
5 7 5
_________
In a division, the divisor is 12 times the quotient and 5 times the remainder. If the remainder is 48, then dividend is ___________
- (a)
240
- (b)
576
- (c)
4800
- (d)
4848
Remainder = 48
\(\therefore\)Divisor = 48 x 5 = 240
Quotient = \(\frac { 240 }{ 12 } \) = 20
Dividend = Divisor x Quotient + Remainder
= 240 x 20 + 48 = 4800 + 48 = 4848.
Which of the following statements is INCORRECT?
- (a)
All even natural numbers which are divisible by 3 are also divisible by 6.
- (b)
If a natural number is divisible by 21, then it is divisible by both 3 and 7.
- (c)
If AB x 4 = 192, then A + B = 10
- (d)
A number of the form 14 N + 2 leaves the remainder 2 when divided by 7
Fill in the blanks
(i) If sum of 3-digit numbers xyz, yzx and zxy is divided by (x + y + z), then quotient is ___P__.
(ii) The difference between 2-digit numbers ab and ba, (where a > b) is divided by 3. The quotient is ___Q___.
(iii) Sum of a 2-digit number and the number obtained by reversing its digits is always divisible by ____R____
- (a)
P Q R 111 3(a + b) 11 - (b)
P Q R 99 (a + b) 7 - (c)
P Q R 111 3(a - b) 11 - (d)
P Q R 99 (a - b) 3
(i) As, xyz = 100x + 10y + z....(i)
yzx = 100y + 10z + x.....(ii)
zxy = 100z + 10x + y.....(iii)
Adding (i), (ii) and (iii), we get
xyz + yzx + zxy
= 100x + 10y + z + 100y + 10z + x + 100z + 10x + Y
= 111x + 111Y + 111z = 111 (x + Y + z)
On dividing by (x + y + z), we get
Quotient = 111.
(ii) ab = 10a + band ba = 10b + a
ab - ba = 10a + b - (10b + a) = 9a - 9b = 9 (a - b)
On dividing by 3, we get
Quotient = 3(a - b)
(iii) Let two digit number be 10x + y.
On reversing the digits, number becomes 10y + x.
Sum = 10x + Y + 10y + x = 11x + 11y = 11 (x + y),
which is always divisible by 11
Match the following.
Column-I | Column-II |
---|---|
(P) If 213x27 is divisible by 9, then x = | (i) 2 |
(Q) If 2415x is divisible by 6 then x = | (ii) 8 |
(R) If 23245x is divisible by 4 and 3, then x = | (iii) 3 |
(S) If 7251x93 is divisible by 11, then x = | (iv) 6 |
- (a)
(P) \(\rightarrow\) (iii); (Q) \(\rightarrow\)(ii); (R) \(\rightarrow\) (iv); (S) \(\rightarrow\) (i)
- (b)
(P) \(\rightarrow\) (ii); (Q) \(\rightarrow\) (iv); (R) \(\rightarrow\) (i); (S) \(\rightarrow\) (iii)
- (c)
(P) \(\rightarrow\) (iii); (Q) \(\rightarrow\) (iv); (R) \(\rightarrow\) (i); (S) \(\rightarrow\) (ii)
- (d)
(P) \(\rightarrow\) (ii); (Q) \(\rightarrow\) (iii); (R) \(\rightarrow\) (i); (S) \(\rightarrow\) (iv)
(P) Since, 213 x 27 is divisible by 9.
So, 2 + 1 + 3 + x + 2 + 7 = 15 + x is divisible by 9.
\(\therefore\)x = 3
(Q) 2415x is divisible by 6, if it is divisible by both 2 and 3.
\(\therefore\)x = 6
(R) 23245x is divisible by 4 and 3
\(\Rightarrow\)5x is divisible by 4
\(\therefore\)Possible values of x are 2, 6
Also, 2 + 3 + 2 + 4 + 5 + x = 16 + x is divisible by 3.
\(\therefore\)Possible values of x is 2.
(S) We have, 7251x 93 is divisible by 11
\(\therefore\) [(7 + 5 + x + 3) - (2 + 1 + 9)] is divisible by 11
\(\Rightarrow\)15 + x - 12 = 3 + x is divisible by 11
\(\therefore\)x can be equal to 8.
How many 5-digit numbers of the form AABAA is divisible by 33?
- (a)
1
- (b)
3
- (c)
0
- (d)
infinite
We have AABAA is divisible by 33.
So, it is divisible by both 3 and 11.
\(\therefore\) A + B + A - (A + A) = B is divisible by 11.
\(\Rightarrow\)B = 0
Also, A + A + B + A + A = 4A + B is divisible by 3.
\(\Rightarrow\)4A is divisible by 3 (.: B = 0)
\(\Rightarrow\)A is divisible by 3
Hence, possible values of A are 0, 3, 6, 9
But A can't be equal to zero.
\(\therefore\)Number of possible 5-digit numbers are 3.
Find the value of A, Band C respectively
(i) A 8 3
x C 9
________
A 0 4 A
+1 5 B B 0
____________
C C A 0 A
___________
(ii) 4 3 A 4
x 3 A
________________
B 7 C 7 6
+ B C 0 C 2 0
_________________
B 4 7 6 9 6
__________________
- (a)
(i) 2 6 7 (ii) 9 5 2 - (b)
(i) 6 7 2 (ii) 4 3 1 - (c)
(ii) 7 5 2 (ii) 9 2 5 - (d)
(i) 7 6 2 (ii) 4 1 3
(i) \(\boxed { 7 } \) 8 3
x \(\boxed { 2 } \) 9
_______________
\(\boxed { 7 } \) 0 4 \(\boxed { 7 } \)
+ 1 5 \(\boxed { 6 } \) \(\boxed { 6 } \) 0
__________________
\(\boxed { 2 } \) \(\boxed { 2 } \) \(\boxed { 7 } \) 0 \(\boxed { 7 } \)
________________
(ii) 4 3 \(\boxed { 4 } \) 4
x 3 \(\boxed { 4 } \)
___________________
\(\boxed { 1 } \) 7 \(\boxed { 3 } \) 7 6
+ \(\boxed { 1 } \) \(\boxed { 3 } \) 0 \(\boxed { 3 } \) 2 0
____________________
\(\boxed { 1 } \) 4 7 6 9 6
____________________