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Practice each question, then open the answer or worked solution.
The sum of a two–digit number and the number obtained by reversing the digits is always divisible by __________.
Solution: 11
Explanation:
Let “a” and “b” be the digits, then X = ab
By reversing the digits, then take Y =ba
Therefore the sum of digits in X is 10a+b
Sum of digits in Y is 10b+a
By adding X and Y, we get
11a+11b = 11(a+b)
Therefore, the number is divisible by 11.
The difference of a two–digit number and the number obtained by reversing its digits is always divisible by ___________.
Solution: 9
Explanation:
Let “a” and “b” be the digits, then X = ab
By reversing the digits, then take Y =ba
Therefore sum of digits in X is 10a+b
Sum of digits in Y is 10b+a
By subtracting X and Y, we get
9a – 9a = 9(ab)
Therefore, the number is divisible by 9.
The difference of a three-digit number and the number obtained by putting the digits in reverse order is always divisible by 9 and
___________.
Solution: 11
Explanation:
Let “a”, “b” and “c” be the digits, then X = abc
By reversing the digits, then take Y =cba
Therefore, the sum of digits in X is 100a+10b+c
Sum of digits in Y is 100c+10b+a
By subtracting X and Y, we get
99(x-z) = (9)(11)(x-z)
Therefore, the number is divisible by 9 and 11.
If
, then A = ___________ and B = ___________
Solution: A = 6 and B= 3
Explanation:
By adding the given digit according to the place value,
20+B + 10 A +B = 80 +A
9A+2B = 60
9A = 60 -2B
As, both A and B are whole numbers, 60-2B should be divisible by 9
So, the value of B should be 3
Substitute B = 3 in 9A + 2B = 60
Simplifying this, we will get
A = 6.
If
, then A = ___________ and B = ___________
Solution: A = 2 and B = 4
Explanation:
From the question, it is given that the units place contains a 6. So, the product of B and B should give a number with 6 in units place. It is possible when B = 4 or 6.
When B = 4,
(10A + B) x (B) = 96
It becomes (10A + 4) x 4 = 96
40A + 16 = 96,
By simplifying the equation, it gives A = 2.
Similarly, when B = 6,
(10A + B)x(B) = 96
By simplifying, it becomes (10A + 6) x 6 = 96
60A + 36 = 96, gives A = 1.
Hence, A = 2 B = 4 or A = 1 B = 6 (Both are feasible solutions)
If
, then B = ___________
Solution: B = 7
Explanation:
From the question, it is clear that the value of B should be greater than 4 and less than 9.
If the B value is less than 4, the solution should have only two digits. If the B value is greater than 9, the solution should have more than 3 digits.
So, the value of B ranges from 4 to 9. The 10’s place, when multiplied by B, gives 49
Hence, B x B = 49
Therefore, B = 7
1 x 35 is divisible by 9 if x = _______.
Solution: 0
Explanation:
We know that, when a number is a multiple of 9, then the sum of digits in the number is divisible by 9
So, by adding the digits, we get:
X+9 = 9
Hence, x= 0
A four-digit number abcd is divisible by 11, if d + b = _______ or _______.
Solution: a+c or b+d
Explanation:
Using the divisibility rule of 11, if abcd is divisible by 11, then
a-b + c-d = 0
By rearranging the terms, we get
a+ c = b+ d, or each should be zero
A number is divisible by 11 if the differences between the sum of digits at its odd places and that of digits at the even places is either 0 or divisible by ______.
Solution: 11
Explanation:
Assume that the four-digit number abcd
Here, the digits at odd places are a and c and in even places, are b and d
Hence, (a+c) – (b+d) = 0
It follows that, (a+c) – (b+d) is divisible by 11, when (a+c) – (b+d) is not equal to 0.
If a 3-digit number abc is divisible by 11, then ______ is either 0 or a multiple of 11.
Solution: (a+c)-b
Explanation:
From the direct consequence from the divisibility rule of 11, we say that (a+c)- b is either 0 or a multiple of 11.