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Problems on Numbers Questions

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7,77,777,7777,77777,777777....
Find last 3 digits of sum of first 26 terms


A452

B456

C732

D672

Answer: Option C

Explanation:

At unit digit 7 would be coming 26 timing
so, 26*7=182.
2 at unit place
Then 25*7=175+ (18 carry)=193
Then 24*7=187
So, Last digits will be 732.,

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How many different 9 digit numbers can be formed from the number 223355888 by re-arranging its digits so that the odd digits occupy even position?


A65 ways

B45 ways

C12 ways

D60 ways

Answer: Option D

Explanation:

Odd places are 4 and these are occupied by 3355. So this can be done in 4!/ (2! 2!) = 6 There are 5 even numbers which have to be placed at 5 odd places. So 5!/(2!3!) = 10 ways so total number of ways of arranging all these numbers are 10 * 6 = 60 ways

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2ab5 is a four digit number divisible by 25. If a number formed from the two digits ab is a multiple of 13, then ab is


A52

B45

C10

D24

Answer: Option A

Explanation:

For a number to be divisible by 25, last two digits of that number should be divisible by 25. So b must be either 2 or 7
It is given that ab must be divisible by 13 and in the options only 52 is divisible by 13.

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When numbers are written in base b we have 12*25=333. The value of b is:


A8

B5

C7

D4

Answer: Option C

Explanation:

Lets assume the base of number = X
Convert all the number to base X
(1*X + 2*X^0)*(2*X + 5*X^0) = (3*X^0 + 3*X^1 + 3*X^2)
By solving the above eq. X=7

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Certain positive integers have these properties:
(I)The sum of the squares of their digits is 50
(II) Each digit is larger than the one to its left.

The product of the digits of the larger integer with both properties is


A7

B36

C25

D48

Answer: Option B

Explanation:

Integers that exhibit the given properties are 17,1236,345
the largest integer among those is 1236
so product of the digits of 1236 is 36.

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A class of 100 students. 24 of them are girls and 32 are not. Which base am I using?


A0

B6

C2

D3

Answer: Option B

Explanation:

Let the base be X.
Therefore (X*X + X*0 + 0) = (2*X +4) + (3*X + 2),
=> X * X = 5 * X + 6,
=> X * X - 5 * X -6 = 0,
=> (X- 6) (X+1) = 0
Therefore base is 6.

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Consider the sequence of numbers 0, 2, 2, 4,... Where for n > 2 the nth term of the sequence is the unit digit of the sum of the previous two terms. Let sn denote the sum of the first n terms of this sequence. What is the smallest value of n for which sn >2771?


A692

B693

C694

D700

Answer: Option B

Explanation:

[0, 2, 2, 4, 6, 0, 6, 6, 2, 8, 0, 8, 8, 6, 4, 0, 4, 4, 8, 2], 0, 2, 2...this series repeats after every 20 terms.


Sum of these 20 terms = 80


So 2771 =34*80 + 51


Sum of 13 terms = 52


So we have to use 34 times 20 terms = 34*20 = 680


680+13 = 693

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Let a, b, c, d and e be distinct integers in ascending order such that(76-a)(76-b)(76-c)(76-d)(76-e) = 1127. What is a + b + c + d


A30

B274

C334

D136

Answer: Option B

Explanation:

Product of 5 terms equal to 1127. As all the five terms are integers, given product should be a product of 5 numbers. Now factorize 1127.
1127 = 72 * 23 = 7 * 7 * 23
But given that all the a, b, c, d, e are distinct. And we are getting only 3 terms with 7 repeats.
Now the logic is, integers means positive and negative, 7 and - 7 possible and 1, - 1 also possible .
As a,b, c, d, e are in ascending order, the factors should be in decreasing order. So (23, 7, 1, -1, -7)
Now a = 53; b = 69; c = 75; d = 77

a + b + c + d = 274.

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A petri dish hosts a healthy colony of bacteria.once a minute every bacterium divides into two.the colony was founded by a single cell at noon.at exactly 12.43(43 minutes later) the petri dish was half full. At what time the dish will be full?


A12.44

B12.54

C12.45

DNone of these

Answer: Option A

Explanation:

Given that the bacteria doubles for every one minute,
After 43 minutes it was half full,
It will become full in the next minute i.e. at 12:44 or after 44 minutes

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A calculator has a key for squaring and one for inverting. So if x is the displayed number, then pressing the square key will replace x by x2 and pressing the invert key will replace x by 1/x. If initially the number displayed is 6 and one alternately presses the invert and square key 16 times each, then the final number displayed (assuming no roundoff or overflow errors) will be


A(6^(-2))^16

B(6^(2))^16

C6^32

D632

Answer: Option B

Explanation:

Even number of inverse key has no effect on the number.
For example, Initially the given number is 6. Square key makes it 6^2 and invert key makes it 1/6^2. Now again square key makes it (1/6^2)^2=1/6^4 and invert key makes it 6^4.
Now observe clearly, after pressing square key 2 times, the power of 6 became 4.
By pressing the square key, the value got increased like 2, 4, 8, .... Which are in the format of 2^n.
So after the 16 pressings the power becomes 2^16
So the final number will be (6^2)^16=6^65536

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