Math Olympiad Test: Real Numbers- 1 - Free MCQ with solutions Class 10

L.C.M. of 144, 180, 192
144 = 2⁴ × 3²
180 = 2² × 3² × 5
192 = 2⁶ × 3
∴ L.C.M. = 2⁶ × 3² × 5 = 2880

Here 445 - 4 = 441; 572 - 5 = 567, 699 - 6 = 693
Now we have to find H.C.F. of 441, 567, 693 is
441 = 3 × 3 × 7 × 7
567 = 3 × 3 × 3 × 3 × 7
693 = 3 × 3 × 7 × 11
The common factors are 3 × 3 × 7 = 63.

Subtract the remainders from the numbers:
398 minus 7 = 391
436 minus 11 = 425
542 minus 15 = 527
Find the highest common factor of 391, 425 and 527. Factorization:
391 = 17 × 23
425 = 17 × 25
527 = 17 × 31
The HCF is 17, so the required number is 17.

In order to find the largest number which divides 615 and 963 leaving remainder 6 in each case, subtract the remainder from both numbers and then find their HCF.
615 minus 6 = 609
963 minus 6 = 957
Prime factorising both 609 and 957 we get,
609 = 3 × 7 × 29 and
957 = 3 × 11 × 29
HCF of 609 and 957 = 3 × 29
= 87
∴ The largest number which divides 615 and 963 leaving remainder 6 in each case is 87.

The smallest number which when divided by 35, 56 and 91 = LCM of 35, 56 and 91
35 = 5 x 7
56 = 2 x 2 x 2 x 7
91 = 7 x 13
LCM = 7 x 5 x 2 x 2 x 2 x 13 = 3640
The smallest number that when divided by 35, 56, 91 leaves a remainder 7 in each case = 3640 + 7 = 3647.

First we find H.C.F. of 408 and 1032
HCF (408,1032) = 24 
Now, As Given
1032m - 408 × 5 = HCF
1032m - 408 × 5 = 24
1032m - 2040 = 24
⇒ 1032m = 24 + 2040
⇒ 1032m = 2064
⇒ m = 2064/1032 = 2
    m = 2

Using the factor tree for prime factorization, we have:

We have 196 = 2² × 7²
2 + 2  = 4

product of number = LCM x HCF
a x b = LCM x HCF
Other number

6ⁿ - 5ⁿ = 6¹ - 5¹ = 1
6² - 5² = 36 - 25 = 11 and so on