Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.

**Answer: **A

Required number = H.C.F. of \((91 - 43)\), \((183 - 91)\) and \((183 - 43)\)

= H.C.F. of \(48, 92\) and \(140 = 4\).

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The greatest four digit number which is divisible by 15, 25, 40, 75 is

**Answer: **B

We want 4 digit number, so option A is not the answer

Now we want greatest numnber, So out of remaining options, 9600 is greatest

9600 is divisible by 25, 75, 40, and 15

So answer is 9600

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Find the H.C.F. of 54, 288, 360

**Answer: **A

Using factorization method,

\(18 = 2 \times 3^2\)

\(288 = 2^5 \times 3^2\)

\(360 = 2^3 \times 3^2 \times 5\)

So H.C.F. will be minimum term present in all three,

i.e. \(2 \times 3^2 = 18\)

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Find the H.C.F. of

\(2^{2}\times 3^{2}\times 7^{2},2\times 3^{4} \times 7\)

**Answer: **B

HCF is Highest common factor, so we need to get the common highest factors among given values.

So we get,

\(2 \times 3 \times 3 \times 7 = 126\)

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The L.C.M. of two numbers is 14560 and their H.C.F. is 13. If one of them is 416, the other is

**Answer: **B

416 X number = 14560 X 13

Therefore, numbr is 455

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A number when divided by 893 the remainder is 193. What will be the remainder when it is divided by 47?

**Answer: **B

Number is divided by 893. Remainder = 193.

Also, we observe that 893 is exactly divisible by 47,

So now simply divide the remainder by 47

Remainder is 5

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The least number which when divided by the numbers 3, 5, 6, 8, 10 and 12 leaves in each case a remainder 2 but which when divided by 13 leaves no remainder?

**Answer: **C

No answer description available for this question.

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The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:

**Answer: **C

Greatest number of 4-digits is 9999.

L.C.M. of 15, 25, 40 and 75 is 600.

On dividing 9999 by 600, the remainder is 399.

Required number,

\((9999 - 399) = 9600\)

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3 birds fly along the circumference of a jungle. They complete one round in 27 minutes, 45 minutes and 63 minutes respectively. Since they start together, when will they meet again at the starting position?

**Answer: **A

We need the instance which means the LCM of times of all 3 birds.

Therefore, LCM = 9 X 3 X 5 X 7 = 945 = 945 minutes

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252 can be expressed as a product of primes as:

**Answer: **A

Clearly, \(252 = 2 \times 2 \times 3 \times 3 \times 7\)

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