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
The H.C.F. takes the lowest power of each shared prime.
\[ 2^{2} \cdot 3^{2} \cdot 7^{2} \quad\text{and}\quad 2 \cdot 3^{4} \cdot 7 \]
\[ \text{H.C.F.} = 2^{1} \cdot 3^{2} \cdot 7^{1} = 126 \]
The L.C.M. would take the highest powers instead.
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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
\[ \text{H.C.F.} \times \text{L.C.M.} = \text{product} \]
\[ 13 \times 14560 = 416 \times n \]
\[ n = \frac{189{,}280}{416} = 455 \]
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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
\[ 893 = 19 \times 47 \]
So \( N = 893k + 193 \) is a multiple of 47 plus 193, and only 193 matters:
\[ 193 = 47 \times 4 + 5 \]
The remainder is 5. This works because 47 divides the original divisor.
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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: B
A number leaving the same remainder 2 under all six divisors is 2 more than a common multiple of them, so start from their lowest common multiple:
\[ \operatorname{LCM}(3,5,6,8,10,12) = 120 \quad\Longrightarrow\quad N = 120k + 2 \]
Now impose divisibility by 13. Since \( 120 = 9(13) + 3 \), we have \( 120 \equiv 3 \pmod{13} \), so
\[ 3k + 2 \equiv 0 \pmod{13} \quad\Longrightarrow\quad 3k \equiv 11 \pmod{13} \]
Multiplying by 9, the inverse of 3 modulo 13 because \( 3 \times 9 = 27 \equiv 1 \):
\[ k \equiv 99 \equiv 8 \pmod{13} \quad\Longrightarrow\quad N = 120(8) + 2 = 962 \]
Check: 962 divided by 3, 5, 6, 8, 10 and 12 leaves 2 every time, and \( 962 = 13 \times 74 \) exactly. Of the distractors, 1562 clears the remainder test but is not a multiple of 13, and 1586 is a multiple of 13 but leaves remainder 1 on division by 5.
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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
They meet at the start again after a whole number of laps each, so take the L.C.M. of the lap times.
\[ 27 = 3^{3}, \quad 45 = 3^{2}\cdot 5, \quad 63 = 3^{2}\cdot 7 \]
\[ \text{L.C.M.} = 3^{3} \cdot 5 \cdot 7 = 945 \text{ minutes} \]
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252 can be expressed as a product of primes as:
Answer: A
Divide out the primes in turn.
\[ 252 = 2 \times 126 = 2 \times 2 \times 63 = 2^{2} \times 3^{2} \times 7 \]
Checking: \( 4 \times 9 \times 7 = 252 \).
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