Aptitude - Numbers - Discussion

Discussion :: Numbers - General Questions (Q.No.30)

30. 

What will be remainder when (6767 + 67) is divided by 68 ?

[A]. 1
[B]. 63
[C]. 66
[D]. 67

Answer: Option C

Explanation:

(xn + 1) will be divisible by (x + 1) only when n is odd.

(6767 + 1) will be divisible by (67 + 1)

(6767 + 1) + 66, when divided by 68 will give 66 as remainder.


Narendra said: (Dec 14, 2010)  
Please suggest me another method.

Shyni M said: (Jan 8, 2011)  
Cant understand the concept .

Nik said: (Jan 31, 2011)  
Any other method please.

Ankush said: (May 31, 2011)  
Simplify please.

Mega said: (Jun 8, 2011)  
Cannot understand the concept please xplain.

Jpragash said: (Jul 17, 2011)  
Not understand.

Mahi said: (Jul 25, 2011)  
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.

Yogesh said: (Jul 27, 2011)  
When x pow even then. what's the procedure?

Deepak said: (Jan 17, 2012)  
Find out unity digit of 67^67....i e unit digit is 3
n 3+67=70
and when divide by 68 remainder 2...ie 68-2=66

Ravi said: (Sep 20, 2012)  
I support deepak but in additional I would like you to know of this 67^67 units digit is to b found out. By cyclicity of numbers concept 67/4 remainder 3. So units digit is units digit of 7^3=**3 so adding 3+67=70.

When divide by 68 remainder 2. Ie 68-2=66.

Padmaja said: (Jun 6, 2013)  
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.

Anuj said: (Sep 16, 2013)  
(67^67+67)/68 = 67^67/68+67/68.

If we 67/68 = -1 remainder.

So ((-1)^67-1)/68 = (-1-1)/68.

i.e.(-2)/68 = remainder can't negative then 68-2 = 66 answer.

Minaxi said: (Oct 18, 2013)  
67((i^67+1)) = 67(1+1) => 67*2 = 134 => 134-68 = 66.

Ramesh said: (May 8, 2014)  
Please give me answer in more better way?

Kalai said: (May 9, 2014)  
We take the odd number 67 and divide it by even 68, the remainder is 1 less than the odd number.

So, the remainder when (67^67 + 67) is divided by 68 will be 67 -1 = 66.

Vijay said: (Jul 20, 2014)  
Check numbers from 3, 4, 5, ... , in the above given format. We will get remainder 1 less than the number, so for 67^67+67 = 66.

Anusha said: (Oct 31, 2014)  
(67^67+67)/68.

--> (67*67*67*...67)/68+(67)/68.

--> 67-68 = -1.

--> (-1*-1*-1*....*-1)/68+(67)/68.

--> (-1/68)+67/68.

--> 68-1= 67.

--> (67+67)/68 = 2.

68-2 = 66.

Nikesh Singh said: (Nov 16, 2014)  
Nothing complexity with the solution.

From the formula : (x^n+1) is always divisible by (x+1) only when n is odd.

In the above problem n is 67 i.e, n is odd.

So, we can write (67^67+67) as ((67 ^ 67)+1) + 66.

Why because making the given expression as convenient to us.

Thank you.

Madhu said: (Dec 6, 2014)  
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?

Anil said: (Feb 8, 2015)  
Just use remainder theorem see how,

Write down the given expression in the form of polynomial of x :67^67+67 = x^67+x, now by remainder theorem (x-(-1)) i.e. 68,

So x^67+x /(x-(-1)) will give you remainder as -2, add this to 68 to get answer 66.

Pooja said: (Mar 25, 2015)  
Please give proper description. Can't understand the problem.

Rishabh said: (Jun 18, 2015)  
67^67 = (68^67-1^67).

So (68^67)+67-1^67.

= (68^67)+66.

= (68^67) is completely divisible by 68. So remainder is 66.

Narendra said: (Jun 21, 2015)  
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.

Prasanna Kartik said: (Jul 8, 2015)  
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.

Prasanna Kartik said: (Jul 8, 2015)  
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.

Abhishek said: (Aug 22, 2015)  
67/68 remainder -1 then (-1) to the power 67 is -1.

+67/68 remainder is 67.

So answer 67-1=66.

Sandeep Jaiswal said: (Sep 24, 2015)  
(67^67+67) = 67 (1^67+1) = 67(1+1) = 134.

134%68 = 66. So answer = 66.

Nitin Sharma said: (Dec 25, 2015)  
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.

Sanjay said: (Jan 12, 2016)  
Simple method of this problem is 67+67 = 134-68 = 66.

Reshma said: (Jul 16, 2016)  
What would be the solution when n is even?

Meghana said: (Jul 27, 2016)  
Why do we subtract 2 the remainder from 68, can anyone explain this?

Nivas said: (Aug 16, 2016)  
The formula is a^n + b^n is divisible by a + b if n is odd.

Ambika said: (Sep 9, 2016)  
Why do we subtract 2 from 68?

Sharanya said: (Sep 17, 2016)  
Give more clarity about this question.

Shru said: (Nov 13, 2016)  
It's very easy, Thank you @Mahi.

Paveek said: (Nov 19, 2016)  
Thank you @Anuj.

Pawas said: (Dec 29, 2016)  
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.

Jitendra Gujjar said: (Dec 30, 2016)  
Is there any Short trick?

Niranjani said: (Feb 6, 2017)  
(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.

Kanak said: (Feb 22, 2017)  
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.

Rudra said: (Aug 15, 2017)  
Trick : the number 67 to the power is given 67 so the power 67 is divisible by 2 remainders comes 1 then 67 of the power is take 1 and then solve i.e.
(67^1+67) divided by 68 = 66.

Siva Ram said: (Sep 13, 2017)  
Take 67 as common 67(1+1)= 134.

Now, divide 134 by 68 you get 66 as remainder, is this easy?

Satheesh Kumar said: (Sep 20, 2017)  
(X pwr n + 1) , if the n value is an even number then what should I do? Can anyone help me?

Vicky said: (Oct 1, 2017)  
Just take the negative remainder so the new equation will be (-1) power 67 +67== (-1+67) ==66.

Pranoti said: (Feb 24, 2018)  
@Kanak.

It was very helpful.

Falandu said: (Jun 10, 2018)  
67 is not odd it is prime.

Lokesh said: (Jul 19, 2018)  
Simple.

For odd = (x^n+1)+(n-1).
For even= (x^n+1)-(n+1).

That's all.

Deepak Parmar said: (Aug 24, 2018)  
How to find unit digit of any number? Please tell me.

Raji said: (Sep 19, 2018)  
(x^n+1) is divisible by (x+1), if not is odd.
So, ( 67^67+1) is divisible by 68 (i.e.)67 + 1.
(67^67+1)÷68 provides remainder 0.

Given: (67^67+67)÷68 ->(67^67 + 1 + 66)÷68.
->provides remainder 66.

Prateeksharma said: (Oct 5, 2018)  
As given (x^n +1) is divisible by (x+1) if and only if n is odd,
In the question we have (67^67 + 67).
So we can write it as( 67^67+ 1+66) = (67^67+1)+66. So that's why we get remainder 66.

Vipin Raj said: (Oct 18, 2018)  
X^n + 1 is divisible by x+1 always whatever the value of n.
X^n -1 is divisible by x+1 and x +1 and x -1.

That same the given value 67^67. +1, +66 extra port of this no will be remainder.

Aakash said: (Jan 13, 2019)  
Please anyone explain clearly. I can't understand.

Mamatha said: (Jul 1, 2019)  
@Sivaram.

It was very helpful. Thanks.

Wesly said: (Jul 24, 2019)  
Thanks for the clear explanation, @Padmaja.

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