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Let n! = 1 x 2 x 3 x……….x n for integer n ≥ 1. If p = 1! + (2 x 2!) + (3 x 3!) + ……(10 x 10!), then p + 2 when divided by 11! Leaves remainder of
If P = 1! = 1
Then P + 2 = 3, when divided by 2! remainder will be 1.
If P = 1! + 2 × 2! = 5
Then, P + 2 = 7 when divided by 3! remainder is still 1.
Hence, P = 1! + (2 × 2!) + (3 × 3!)+ ……+ (10 × 10!)
Hence, when p + 2 is divided by 11!, the remainder is 1.
Alternative method:
P = 1 + 2 × 2! + 3 × 3! + …..10 × 10!
= (2 – 1)1! + (3 – 1)2! + (4 – 1)3! + …+ (11 – 1)10!
= 2! – 1! + 3! – 2! + …..+ 11! – 10!
= 11! – 1
Hence p + 2 = 11! + 1
Hence, when p + 2 is divided by 11!, the remainder is 1
The remainder, when (15^{23} + 23^{23}) is divided by 19, is:
a^{n }+b^{n} is always divisible by a + b when n is odd.
Therefore 15^{23} + 23^{23} is always divisible by 15 + 23 = 38.
As 38 is a multiple of 19, 15^{23} + 23^{23} is divisible by 19.
Therefore,the required remainder is 0.
What is the sum of all twodigit numbers that give a remainder of 3 when they are divided by 7?
First of all, we have to identify such 2 digit numbers.
Obviously, they are 10, 17, 24, ….94
The required sum = 10 + 17 … 94.
Now this is an A.P. with a = 10, n = 13 and d = 7
Hence, the sum is
After the division of a number successively by 3, 4 and 7, the remainders obtained are 2,1 and 4 respectively. What will be the remainder if 84 divides the same number?
In the successive division, the quotient of first division becomes the dividend of the second division and so on.
Let the last quotient be p, so the last dividend will be 7p + 4 which is the quotient of the second division.
So, the second dividend is (7p + 4) × 4 + 1.
Applying the same logic, the number = 3 {4(7p + 4) + 1} + 2 = 84p + 53
Hence, if the number is divided by 84, the remainder is 53.
Let N = 55^{3} + 17^{3} – 72^{3}. N is divisible by:
We have N = 55^{3} + 17^{3} – 72^{3} = (54 + 1)^{3} + (18 – 1)^{3} – 72^{3}
When N is divided by 3, we get remainders (1)^{3} + ( 1)^{3} – 0 = 0
Hence, the number N is divisible by 3.
Again N = (51 + 4)^{3} + 17^{3} – (68 + 4)^{3}
When N is divided by 17, the remainder is (4)^{3} + 0 – (4)^{3} = 0
Hence, the number is divisible by 17.
Hence, the number is divisible by both 3 and 17.
What will be the unit digit of 13^{41 }?
As we know the last digit depends upon the unit digit of the multiplier numbers so the unit digit of 13^{41}
is same as the last digit of 3^{41} and we know that the cyclicity of 3 is 4
On dividing the number 41 with 4 and we will get the remainder as 1 and the last digit will be 3^{1} = 3
What will be the unit digit when 45^{45}
Here the last digit is depend upon the 5
The cyclicity of 5 is 1 so there is no need to divide the power with 1
because whatever is the power of 5 the last digit will remains 5 so the last
digit of 45^{45} will be 5 .
5^{1} = 5
5^{2} = 25
5^{3 }= 125
What will be the last digit of 3^{4 }x 4^{5} x 5^{6}
Last digit of 3^{4 }= 1
Last digit of 4^{5} = 4
Last digit of 5^{6 }= 5
So as we discussed earlier that the last digit depends upon the last digits
So 1 x 4 x 5 = 20 so the last digit is 0; also whenever 5 is multiplied by even
number then last digit will be 0
Identify the last digit of (79^{4 }+ 87^{5})
Last digit of 79^{4} is depends upon the the 9^{4} i.e 1
Last digit of 87^{5} is depends upon the the 7^{5}i.e 7
So the last digit of the whole expression will be 1 +7 = 8
So the last digit will be 8
What will be the last digit of 43^{56} x 5^{67 }x 45^{34}
Last digit of 43^{56 }= 1 because cyclicity of 3 is 4 and 56 is completely
divisible by 4 so the last term of 3^{4 }is 1.
Last digit of 5^{67} is 5 and last digit of 45^{34} is also 5
so that means the last digit of whole of the expression is 1 x 5 x 5 = 5 therefore the digit is 5.
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