NIMCET 2007 — Reasoning PYQ

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Question

NIMCET 2007 — Reasoning PYQ

NIMCET | Reasoning | 2007

The smallest number which when divided by 2, 3, 4, 5 leaves 1 as remainder in each case but, leaves no remainder when divided by 7:

Choose the correct answer:

Explanation

Step 1: Find the Least Common Multiple (LCM)

To find a number that is divided by $2, 3, 4,$ and $5$ leaving a remainder of $1$ , we first need to find the smallest number completely divisible by them. This is their LCM.

Prime factorization:
- $2 = 2^{1}$
- $3 = 3^{1}$
- $4 = 2^{2}$
- $5 = 5^{1}$

$LCM (2, 3, 4, 5) = 2^{2} \times 3^{1} \times 5^{1} = 4 \times 3 \times 5 = 60$

Step 2: Express the Required Number in General Form

Any number that leaves a remainder of $1$ when divided by $2, 3, 4,$ and $5$ can be written in the mathematical form:

$Number = 60 k + 1$

(where $k$ is a positive integer: $k = 1, 2, 3, \dots$ )

Step 3: Apply the Divisibility Condition for 7

The question states that the number must leave no remainder when divided by $7$ (i.e., it must be perfectly divisible by $7$ ).

Let's simplify the expression $60 k + 1$ by breaking down $60$ into a multiple of $7$ :

$60 k + 1 = (56 k + 4 k) + 1 = 56 k + (4 k + 1)$

Since $56 k$ is always perfectly divisible by $7$ , we only need to find the smallest value of $k$ for which $(4 k + 1)$ is divisible by $7$ .

For $k = 1$ : $4 (1) + 1 = 5$ (Not divisible by 7)
For $k = 2$ : $4 (2) + 1 = 9$ (Not divisible by 7)
For $k = 3$ : $4 (3) + 1 = 13$ (Not divisible by 7)
For $k = 4$ : $4 (4) + 1 = 17$ (Not divisible by 7)
For $k = 5$ : $4 (5) + 1 = 21$ (Divisible by 7 because $21 \div 7 = 3$ )

Step 4: Calculate the Smallest Number

Substitute $k = 5$ back into our general form equation:

$Number = 60 (5) + 1$

$Number = 300 + 1 = 301$