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Examples 21 to 25 of Fundamental(Numbers)
- August 18, 2020
- Category: Examples 21 to 25 of Fundamental(Numbers)
(21) What least number must be added and subtract to 3000 to obtain a number exactly divisible by 19 ?
Soln. On dividing 3000 by 19, we get 17 as a remainder.
∴ Number to be added
= ( 19 – 17 )
= 2
& Number to be subtracted
= 17
(22) Find the number which is nearest to 3115 and is exactly divisible by 23.
Soln. On dividing 3105 by 23, we get 10 as a remainder.
∴ Number to be added
= ( 23 – 10 )
= 13
& Number to be subtracted
= 10
Hence, required nearest number = 3115 – 10 = 3105.
(23) A number when divided by 437 gives a remainder 47. When the same number is divided by 19, what would be the remainder ?
Soln.
The easiest value of
b is 0.
∴ We get,
Number = 47
∴ On dividing 47 by 19 we get the remainder 9.
(24) A number being successively divided by 3, 5 and 8 leaves remainders 1, 4 and 7 respectively. Find the respective remainders if the order of divisors be reversed.
Soln.
Soln. Let d = 0, then c
= 8 × 0 + 7
= 7
Now, b
= 5 × 7 + 4 = 39
And a
= 3 × 39 + 1 = 118
Hence, that number = 118. & Now,
∴ Respective remainders are 6, 4, 2.
(25) How many numbers between 20 and 100 are divisible by 7 ?
Soln. The required numbers are 21, 28, 35, …, 98.
∴ = 7n + 14 ⇒ 7n + 14 = 98
∴ n = 12