Find the greatest number that will divide 400, 435 and 541 leaving 9, 10 and 14 as remainders respectively
Answer: B
Subtract each remainder, then take the H.C.F. of the results.
\[ 400 - 9 = 391, \quad 435 - 10 = 425, \quad 541 - 14 = 527 \]
\[ 391 = 17 \times 23, \quad 425 = 17 \times 25, \quad 527 = 17 \times 31 \]
\[ \text{H.C.F.} = 17 \]
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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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The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:
Answer: D
The number is 8 more than a common multiple of all four.
\[ 12 = 2^{2}\cdot 3, \ 15 = 3\cdot 5, \ 20 = 2^{2}\cdot 5, \ 54 = 2\cdot 3^{3} \]
\[ \text{L.C.M.} = 2^{2} \cdot 3^{3} \cdot 5 = 540 \]
\[ 540 + 8 = 548 \]
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Find the largest number which divides 62,132,237 to leave the same reminder.
Answer: C
When the remainder is the same but unknown, take the H.C.F. of the differences - the remainder cancels out of each.
\[ 132 - 62 = 70, \quad 237 - 132 = 105, \quad 237 - 62 = 175 \]
\[ \text{H.C.F.}(70, 105, 175) = 35 \]
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Find the H.C.F. of \(\dfrac{2}{3}, \dfrac{4}{6}, \dfrac{8}{27}\)
Answer: A
Formula,
\(H.C.F. = \dfrac{H.C.F. of Numerators}{L.C.M. of Denominators}\)
So,
\(\dfrac{H.C.F. of (2,4,8) }{L.C.M. of (3,6,27)} \\~\\=\dfrac{2}{27}\)
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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 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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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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Rajesh had to arrange his books in uniform groups. He makes groups of 4 books each. But 3 books are left. He tries it with groups of 5 books each. But still 3 books are left. 3 books are still left when he tried with groups of 9 or 10 books each. How many books does he have?
Answer: D
Three books are left over in every arrangement, so the total is 3 more than a common multiple of all four group sizes:
\[ N = \operatorname{LCM}(4,5,9,10)\,k + 3 \]
Taking prime factors, \( 4 = 2^{2} \), \( 5 = 5 \), \( 9 = 3^{2} \), \( 10 = 2 \cdot 5 \):
\[ \operatorname{LCM} = 2^{2} \cdot 3^{2} \cdot 5 = 180 \]
\[ \therefore\; N = 180k + 3, \quad\text{smallest such } N = 183 \]
The other options are all multiples of the LCM with nothing left over, so each would divide exactly and leave no remainder at all.
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If the product of two numbers is 84942 and their H.C.F. is 33, find their L.C.M.
Answer: A
\(H.C.F. \times L.C.M. = 84942\), because we know
Product of two numbers = Product of H.C.F. and L.C.M.
\(L.C.M. = 84942 \div 33\)
\(= 2574\)
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