The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
Answer: D
L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be \(90k + 4\), which is multiple of \(7\).
Least value of k for which \((90k + 4)\) is divisible by \(7\) is \(k = 4.\)
Required number,
\(= (90 \times 4) + 4\)
\(= 364\)
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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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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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Find the H.C.F. of 2.1, 1.05 and 0.63.
Answer: C
Clear the decimals by scaling all three by 100, then rescale the answer.
\[ \text{H.C.F.}(210, 105, 63) = 21 \]
\[ \frac{21}{100} = 0.21 \]
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The Greatest Common Divisor for two numbers is 8 while their Least Common Multiple is 144. Find the other number if one number is 16.
Answer: C
For any two numbers,
\[ \text{H.C.F.} \times \text{L.C.M.} = \text{product of the numbers} \]
\[ 8 \times 144 = 16 \times n \quad\Longrightarrow\quad n = \frac{1152}{16} = 72 \]
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If the LCM of two numbers is 138. But their GCD is 23. The numbers are in a ratio 1:6. Which is the largest number amongst the two?
Answer: B
With H.C.F. 23 the numbers are \( 23m \) and \( 23n \) with \( m \) and \( n \) coprime, and the ratio \( 1 : 6 \) makes them 23 and 138.
\[ \text{L.C.M.} = 23 \times 1 \times 6 = 138 \quad\checkmark \]
So the larger number is 138.
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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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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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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 sum and HCF of two numbers are 156 and 13. The numbers of such number pairs is
Answer: A
If the HCF is 13, both numbers are 13 times something, and those somethings share no factor. Write the pair as \( 13a \) and \( 13b \) with \( \gcd(a,b)=1 \):
\[ 13a + 13b = 156 \quad\Longrightarrow\quad a + b = 12 \]
So the question becomes: how many coprime pairs sum to 12?
Two pairs qualify. The coprimality condition is the whole point: without it the HCF would be larger than 13 and the given data would be contradicted.
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