**What is the remainder when 77+777+7777+77777.........+777777777 is divided by 8?**

__Question:__**Options:**

A. | -1 | |

B. | 1 | |

C. | 7 | |

D. | 0 | |

E. | Cannot be determined |

**Answer: Option D**

: D

78 remainder = -1 Thus, any odd power of 7, when divided by 8, will give a remainder of -1 itself. (−1)7+(−1)77+(−1)777......8 = -1-1-1-...8 times = - 8 The remainder when -8 is divided by 8, is 0.

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## More Questions on This Topic :

**An elevator starts at the basement with 9 people (excluding the elevator operator) and discharges them all by the time it reaches the top floor, which is floor number 6. In now many ways could the operator have perceived the people leaving the elevator if all the people look the same? (there are totally 6 floors excluding the basement)**

__Question 1.__- 2002
- 1009
- 1576
- 2492
- Cannot be determined

**Answer: Option A**

: A

Thisis a question based on distributing similar things to different groups (SàD) i.e. the distribution of 9 similar things to 6 different groups A+B+C+D+E+F=9 Solving, we get 14C5= 2002.

**Three distinct prime numbers less than 10 are taken and all the numbers are formed by arranging all the digits taken. Now the difference between the largest and the smallest number is 495. It is also given that the sum of the digits is more than 13. What is the product of the digits?**

__Question 2.__- 42
- 70
- 105
- Cannot be determined
- 5222

**Answer: Option B**

: B

Prime numbers less than 10 = 2, 3, 5, 7. If the difference between the largest and the smallest number is ending in 5, the prime numbers in the end position have to be 7 and 2. The smallest and largest numbers are of form 2_7 and 7_2

Since it is given that the sum of the digits is >13, x will be 5.

Verifying, 752-257 = 495. Answer is option (b).

**How many pairs of positive integers, x, y exist; such that x2+3y and y2+3x are both perfect squares?**

__Question 4.__**Answer: Option A**

: D

Since x and y are positive, we may write - x2+3y=(x+a)2, and

y2+3x=(y+b)2 where a, b are positive integers. Expanding, we find that the squared terms cancel, leaving the linear simultaneous equations -

3y=2ax+a2

3x=2by+b2

Solving, we obtain -

x=2a2b+3b29−4ab

y=2b2a+3a29−4ab Since a and b are positive, the numerator in each fraction will be positive. For the denominator to be positive, we must have ab = 1 or 2. If (a,b) = (1,1), (1,2), (2,1), then, respectively, (x,y) = (1,1), (16,11), (11,16). Hence, these are the only solutions.

**If S1,S2,S3,....Sn are the sums of infinite geometric series whose first terms are 1, 2, 3,...n and whose ratios are 12,13,14,1(n+1) respectively, then find the value of S21+S22+S23+....+S26**

__Question 6.__- 140
- 139
- 91
- 140
- 5222

**Answer: Option B**

: B

S1=11−12=2

S2=21−13=3

Thus, the respective sums are 2,3,4,5,6,7

Sums of square of these numbers =[n(n+1)(2n+1)6]−1=139.

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