What is the remainder when [7^{(4n + 3)}]*6^{n} is divided by 10; where ‘n’ is a positive integer.
[7^{(4n + 3)}]6^{n}
= 7^{4n} x 7^{3} x 6^{n}
= 49^{2n} x 7^{3} x 6^{n}
= When each factor is divided by 10 the remainders in each case = (1)^{2n}, 3, and 6 (6 when raised to the power of any natural number is divided by 10 always gives remainder as 6 itself)
So, all the remainders thus found above are 1, 3 and 6
So their multiplication= 1*3*6= 18
So the remainder after 18 has been divided by 10 = 8 (option ‘C’)
Numbers 11284 and 7655, when divided by a certain number of three digits, leave the same remainder. Find that number of 3 digits and their sum.
One has to remember that each factor of the difference of two numbers gives the same remainder if those numbers are divided by it.
Now the difference here = 11284 – 7655 = 3629
Factors of 3629 are 1, 19, 191 and 3629
But we have to find the three digit number here, so 191 is the required number and their sum is 1+9+1 = 11 (option ‘A’)
64329 is divided by a certain number. While dividing, the numbers 175, 114 and 213 appear as three successive remainders, the divisor is?
We have three remainders, means the number comprising of the first digits i.e. 643 was divided first and we got 175 as the remainder.
Now according to DIVIDEND = DIVISOR x QUOTIENT + REMAINDER
=> DIVISOR x QUOTIENT = DIVIDEND – REMAINDER
=> DIVISOR x QUOTIENT = 643 – 175 = 468
We see that 468 is divisible by 234 only among all the answer options; so 234 (option ‘C’) is the divisor we need.
Find smallest number that leaves remainder 3, 5, 7 when divided by 4, 6, 8 respectively.
See carefully the difference between the divisor and remainder is having a certain trend. i.e.4 – 3 = 6 – 5 = 8 – 7=1
In such questions, take LCM of divisors and subtract the common difference from it.Now the LCM of 4, 6, 8 = 24Therefore the required number here = 24 – 1 = 23 (option ‘A’)
Find smallest number that leaves remainder 3, 4, 5 when divided by 5, 6, 7 respectively and leaves remainder 1 when divided by 11.
We have just seen above in TYPE2 how to tackle the first part of the questionThus the number for the first part would be the [(LCM of 5, 6, 7) – (Common difference of divisors and their remainders)] i.e. 210 – 2 = 208
Here now, we have one more condition to satisfy i.e. remainder 1 when divided by 11
Here we should remember that if LCM of the divisors is added to a number; the corresponding remainders do not change i.e if we keep adding 210 to 208… the first 3 conditions for remainders will continue to be fulfilled.
Therefore now, let 208 + 210k be the number that will satisfy the 4th condition i.e. remainder 1 when (208 + 210k)/11
Now let’s see how
The expression (208 + 210k)/11 = 208/11 + 210k/11
Now the remainder when 208 is divided by 11 = 10
And remainder when 210k is divided by 11 = 1*k = k
Therefore the sum of both the remainders i.e. 10 + k should leave remainder 1 on division of the number by 11
Obviously k = 2
Hence the number = 208 + 210*2 = 628 (option ‘D’)
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