Question
The product of two numbers is 3600 and their HCF is 6.
Total number of such pairs of numbers is?Solution
As HCF is 6 so let numbers be 6x & 6y. It means x & y must be co-prime. Now 6x × 6y = 3600, So xy = 100. Now pairs of (x, y) can be (1,100) , (2,50) , (4, 25) , (5, 20) & (10, 10). Here only (1,100) & (4, 25) are co-prime so number of pairs are 2.
More HCF and LCM Questions
Determine the simplified value of the expression: 12 × 15 - 20 + 15 + 12 - 18 + 3 × 4 + 18.
12.5% of 45 % of 480 + 957/29 = ? -14 × 15
22 + 60 × 3 ÷ 12 = ?
(√121 + √196) × 7 =? × 5
Solve the following:
240 ÷ 4 × 512 ÷ 8
45% of 360 - 160 + ? = √324
[192 ÷ 6 × 5] ÷ (? + 3) = 20
What will come in place of (?) in the given expression.
{(60% of 250) + √144} ÷ (3² - 4) = ?1.25 × 36 + 2.75 × 40 = ? × 3.1
60 % of 640 - 57 × 2 - 1520 / 38 = ?
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