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

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 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 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. |

Random Retard said: (Dec 18, 2019) | |

Thanks to all the good souls who took time to explain the answer. |

Vasanthi said: (Jan 22, 2020) | |

I didn't understand the concept. Please explain it for me. |

Kamal Hachan said: (Mar 23, 2020) | |

(67^67+67)%68 where % is mod operator. it gives remainder. = 67^67%68+67%68 =(68-1)^67%68 +(68-1)%68 =(-1)^67+(-1) = -1-1 =-2 (Since remainder can't be negative ) =68-2 = 66 believe me this is best trick for remainder If any doubt please email me. |

Bhaskar said: (May 26, 2020) | |

By using binomial theorem. ((68-1)^67+67)/68. All term which includes 68 will be divided by 68. Then((-1)^67+67)/68 will be left, It becomes 66/68. Hence remainder is 66. |

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