Question

Find the natural numbers between 101 and 999 which are divisible by both 2 and 5. [CBSE 2014]

Answer 1

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THE NO. WHICH ARE DIVISIBLE BY 2 AND 5
ARE ABSOLUTELY DIVISIBLE BY 10

SO FOR FINDING U SHOULD START TO SEARCH THE NO. DIVISIBLE BY 10

SO D= 100(DIFFERENCE)

FIRST TERM = 110 = a

LAST TERM = 990 = Tn

NOW

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Tn = a + d ( n-1)

990 = 110 + 10 ( n-1 )

990 = 110 + 10n -10

990 = 100 + 10n

10n = 990 - 100

10 n = 890

n = 890 / 10

n = 89

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Answer 2

For a number to be divisible by both 2 and 5

⇒ The number must be divisible by 2 x 5 = 10

Find the first number:

Smallest number in the series that is divisible by 10 is 110

Find the last number:

Largest number in the series that is divisible by 10 is 990

Form the nth term for the AP:

an = a1 + (n - 1)d

⇒ a1 = first term = 110

⇒ d = common difference = 10

an = 110 + (n - 1) (10)

an = 110 + 10n - 10

an = 100 + 10n

Find the number of terms:

an = 100 + 10n

990 = 100 + 10n

10n = 990 - 100

10n = 890

n = 890 ÷ 10

n = 89

Answer: There are 89 terms.