Are number was divided successively in order by 4 5 and 6 the remainders were respectively 2 3 and 4 the number is?

Given:

The number, when divided by 4, 5 and 7, leaves remainder 2, 3 and 4 respectively.

Concept Used:

(Divisor × Quotient) + Remainder = Dividend

Concept of successive division

Calculation:

Here,

d1 = 4, d2 = 5 and d3 = 7

 r1 = 2, r2 = 3 and r3 = 4

Start from last digits

The number gives a remainder of 4 when it is divided by 7

⇒ (7x + 4)
The number gives remainder 3 when it is divided 5

⇒ [5 × (7x + 4) + 3]
Also, the number when divided by 4 gives remainder 2

⇒ [4 × {5 × (7x + 4) + 3} + 2]

⇒ [4 × (35x + 20 + 3) + 2]

⇒ 140x + 94  [∵ Least quotient is 1]

⇒ 140 + 94 = 234

∴ Such least number is 234

 If the quotient of the dividend is taken and this is used as the dividend in the next division, such as division is called Successive division. A successive division process can continue up to any number of steps until the quotient in a division became zero for the first time, i.e., the quotient in the first division is taken and divided in the second division; the quotient in the second division is taken as the dividend in the third division; the quotient in the third division is taken as the dividend in the fourth division and so on.

We will have to follow the rule of Successive Division (Not Normal Division). Hence, LCM Method is not followed there.

 Alternate Method

We can Multiply 7 with 1 and add 4 (7 × 1 + 4) = 11

then, multiply 11 with 5 and add 3 (11 × 5 + 3) = 58

then, multiply 58 with 4 and add 2, we will get (58 × 4 + 2) = 234

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