Aptitude - Numbers - Discussion

a^b is a raise to the power b.

To understand this question we must take a simpler example.
for ex- (3x3)/4

3 = (4-1),
Therefore the question becomes
(4-1)(4-1)/4.
On multiplying (4^2-4-4+1)/4
Note that all the terms in the expansion are completely divisible by 4 except 1.hence Remainder will be 1.

Now take expression for example- 3x3x3 /4.
again (4-1)(4-1)(4-1)/4,
(4^2-4-4+1)(4-1) / 4,
= 4x(4^2-4-4+1)-1x(4^2-4-4+1),
= 4x(4^2-4-4+1) + (-4^2)+4+4-1.

Here the first term is divisible by 4 and in the second term after multiplication by -1 only +1 remains which is not divisible.

Carrying out this process only 1 will remain as the remainder and its sign will depend on the power of the numerator. when the power of 3 was 2, the remainder was +1 and when the power was 3, the remainder was -1.

This means when the power of numerator is odd, -1 will remain and when the power will be even, +1 will remain.

Using this in our example:- (67^67)/68.

-1 will remain because the power of the numerator is odd.
and we also have a +67 in the numerator. therefore, total remaining in the numerator is -1+67 = 66. Which is the remainder and our answer.

Take two simpler numbers, like 2 and 2+1=3.

If we take (2^2 + 2)/3, that is seen to be 6/3, so the remainder is 0.
Is it 0? Try the next number.

If we take (3^3+3)/4, that is seen to be (27+3)/4 = 30/4, so the remainder is 2.

This one is 1/2, so it is not always 0.

If we take (4^4 + 4)/5, we get (256 + 4)/5 = 260/5 = 0.
Now from this one, make it is 0 for even's, 2 for odds.

If we take (5^5 + 5)/6, we get (3125+5)/6 = 3130/6 = 521, remainder 4.
Since we have an odd, we'll keep going with odds. So far we have the remainder is 1 less than the starting number.

If we take (7^7 + 7)/8, we get (823,543+7)/8 = 823,550/9 = 102943 remainder 6.

Again, we get a remainder of 6, which is 7-1.

To try the next odd, take (9^9-9)/10 = (387,420,489+9)/10 = 38,420,498/10, which has a remainder of 8, which is, again one less than what we started with.

Given this, we known that (67^67 + 67) mod 68 is 67-1 = 66.

22 + 2 divided by 3.
= 4 + 2,
= 6.
so when 6 divided by 3 then remainder = 0.

So by doing calculations like this we can get:

22 + 2 divided by 3 then remainder = 0.
33 + 3 divided by 4 then remainder = 2.
44 + 4 divided by 5 then remainder = 0.
55 + 5 divided by 6 then remainder = 4.

So by observing above examples we can say;

xx + x is divided by (x+1) then the remainder is (x-1) where x is odd number.

So, now we can say when 6767 + 67 is divided by 68 then remainder is 66.

It's quite simple. Find unit digit of power value add it with their next value. Divide with divisor. Then subtract remainder from original divisor.

For ex: (3^3+3)/4.

Unit digit of 3^3 is 7. Add with 3 as it is given 7+3=10.

Divide 10/4 remainder=2.

Now finally subtract 2 with 4 so 4-2=2.

Let's see this one: (67^67+67)/68.

Unit digit of 67^67 is 3 as (7^16*7^3). Add this with 67.67+3 = 70.

Divide 70/68 = remainder = 2.

Finally subtract 2 with 68 so 68-2 = 66 answer.

Simply if you got dis situation do as follow

jus 67*2=134

68)134(1
   68
   ---
   66

if u hav any doubt check wid another simple example as follow
3pwr3+3 check wedr it is divisible with 4 r not

by solving in normal way we will get as follow
3pwr3+3 you will get 30. divide 30 with 4 you will get 2 reminder


according to our procedure
3*2=6

4)6(1
  4
  --
   2

So better to go this way guys.

If (67^67 + 1) + 66, when divided by 68 will give 66 as remainder.

Again if you use 63 instead of 67 (67^67 + 1) + 63, when divided by 68 will give 63 as remainder.

Same goes with (67^67 + 1) + 67, when divided by 68 will give 67 as remainder.

WHAT LOGIC IS THERE? AND THE QUESTION IS DIRECTLY MENTION THAT (67^67+1) +67 IS DIVIDE BY 68.

FIND THE REMAINDER THEN WHY ARE WE ADDING EXTRA NUMBERS ON IT AGAIN?

Dividend = Divisor*Quot+Remainder.

Here, Dividend is '67^67 + 67' and divisor is 68.

i.e 67^67 + 67 = 68*Quot+Remainder.....(1).

Now, we know that (x^n + 1) will be divisible by (x + 1) only when n is odd. --> Theory.

We express Dividend as follows: (67^67+1)*1+66.....(2).

i.e. 67^67+67 = 68*Quot+Remainder.....(from 1).

Comparing the above.

(67^67+67)/68 = 67^67/68 + 67/68.
= (67+67)/68.

a^n/(a+1) = a when n is odd.
= 66(ans),
= 1 when n is even.

Hi guys,

Here you can use concept of negative reminders.

68/67 here positive reminder as all we know 68, but negative reminder is -1(67-68).

For example 11/3 positive reminder 2 and negative reminder -1(11-12).

Here is our question.

Rem of (67^67+67)/68 = (-1)^67+(-1) = -2.

So the answer is 68-2 = 66.

Hi guys,

Here you can use concept of negative reminders.

68/67 here positive reminder as all we know 68, but negative reminder is -1(67-68).

For example 11/3 positive reminder 2 and negative reminder -1(11-12).

Here is our question.

Rem of (67^67+67)/68 = (-1)^67+(-1) = -2.

So the answer is 68-2 = 66.