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# What is the sum of all two digit numbers which leave a remainder of 6 when divided by 8?(CAT, 2012)

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**Solution:**

**What is the sum of all two digit numbers which leave a remainder of 6**

**Before we solve, let’s understand that division is equivalent to subtraction.**

**This means when we divide we actually subtract the divisor from the dividend that number of times as the quotient.**

**For example \frac {10}{2} = 5, this means subtract 2 from 10 five times only, if the dividend is divisible completely as in our case.**

**So, when a question is saying divide by 8, it is actually saying subtract 8 from the two digits number.**

**For creating remainder, what we do that we add the required remainder to the number, dividend, completely divisible by a divisor.**

**For example \frac {16}{8} = 2, but to have a remainder 6, we add 6 to 16 and it now becomes 8+6=14.**

**We know that the two digits numbers range from 10 to 99**

**Before 10, we have 9 which is having only 1 digit; and after 99, we have 100 which have 3 digits.**

**The first two digit number in our case is 14**

**This means we are actually having an Arithmetic sequence or Arithmetic Progression of the following form:**

**What is an Arithmetic sequence? Click here.**

**14, 22, 30…94**

**So the first number { a }_{ 1 } is = 14**

**Common difference (c.d) is = 8**

**To find the number of terms in this A.p (learn how to derive the formula here)**

**{ a }_{ L }={ a }_{ n }+(n-1)c.d**

** 94= 14+(n-1)8**

**94= 14+8n-8**

**94=14-8+8n**

**94=6+8n**

**94-6=8n**

**\frac { 88 }{ 8 } = 11 terms**

**To find the sum of this A.P (learn how to derive the formula here)**

** { S }_{ n }=\frac { n[2a+(n-1)c.d] }{ 2 } = \frac { n }{ 2 } [2a+(n-1)c.d]**

** { S }_{ 11 }=\frac { 11 }{ 2 } [2\times 14+(11-1)8]**

** { S }_{ 11}=\frac { 11 }{ 2 } [2\times 14+80]**

** { S }_{ 11 }=\frac { 11 }{ 2 } [28+80]**

** { S }_{ 11 }=\frac { 11 }{ 2 } \times 108**

**11x 54= 594**

**Hence the option (b) is correct.**

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3,180 total views, 2 views today

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