The least number which when divided by 16,18,20 and 25 leaves 4 as remainder in each case but when divided by 7 leaves no remainder is

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17004

18004

18002

18000

answer is D.

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Detailed Solution

To solve this problem, we need to find the least number $N$ $N$ that satisfies the following conditions:

When divided by 16, 18, 20, and 25, the remainder is 4.
When divided by 7, there is no remainder (i.e., $N$ $N$ is divisible by 7).

Step 1: Express the conditions mathematically

The number $N$ $N$ leaves a remainder of 4 when divided by 16, 18, 20, and 25. This means that: $N \equiv 4 \ (\text{mod} \ 16)$ $N$ $\equiv$ $4$ $($ $mod$ $16$ $)$ $N \equiv 4 \ (\text{mod} \ 18)$ $N$ $\equiv$ $4$ $($ $mod$ $18$ $)$ $N \equiv 4 \ (\text{mod} \ 20)$ $N$ $\equiv$ $4$ $($ $mod$ $20$ $)$ $N \equiv 4 \ (\text{mod} \ 25)$ $N$ $\equiv$ $4$ $($ $mod$ $25$ $)$ These conditions are equivalent to: $N - 4 \equiv 0 \ (\text{mod} \ 16), \ (\text{mod} \ 18), \ (\text{mod} \ 20), \ (\text{mod} \ 25)$ $N$ $-$ $4$ $\equiv$ $0$ $($ $mod$ $16$ $)$ $,$ $($ $mod$ $18$ $)$ $,$ $($ $mod$ $20$ $)$ $,$ $($ $mod$ $25$ $)$ So $N - 4$ $N$ $-$ $4$ must be divisible by the least common multiple (LCM) of 16, 18, 20, and 25.

Step 2: Calculate the LCM of 16, 18, 20, and 25

To find the LCM, first, break down the numbers into their prime factorizations:

16 = $2^4$ $2$ $4$
18 = $2 \times 3^2$ $2$ $\times$ $3$ $2$
20 = $2^2 \times 5$ $2$ $2$ $\times$ $5$
25 = $5^2$ $5$ $2$

The LCM is obtained by taking the highest powers of all prime factors:

\text{LCM}(16, 18, 20, 25) = 2^4 \times 3^2 \times 5^2 = 16 \times 9 \times 25 = 3600

$LCM$ $($ $16$ $,$ $18$ $,$ $20$ $,$ $25$ $)$ $=$ $2$ $4$ $\times$ $3$ $2$ $\times$ $5$ $2$ $=$ $16$ $\times$ $9$ $\times$ $25$ $=$ $3600$

So, $N - 4$ $N$ $-$ $4$ must be divisible by 3600. Therefore, $N = 3600k + 4$ $N$ $=$ $3600$ $k$ $+$ $4$ , where $k$ $k$ is an integer.

Step 3: Apply the condition that $N$ $N$ is divisible by 7

Now, we need to find the value of $N$ $N$ that is divisible by 7. Thus:

3600k + 4 \equiv 0 \ (\text{mod} \ 7)

$3600$ $k$ $+$ $4$ $\equiv$ $0$ $($ $mod$ $7$ $)$

We first find $3600 \mod 7$ $3600$ $mod$ $7$ :

3600 \div 7 = 514 \text{ remainder } 2

$3600$ $\div$ $7$ $=$ $514$ $remainder$ $2$

So, $3600 \equiv 2 \ (\text{mod} \ 7)$ $3600$ $\equiv$ $2$ $($ $mod$ $7$ $)$ , and the equation becomes:

2k + 4 \equiv 0 \ (\text{mod} \ 7)

$2$ $k$ $+$ $4$ $\equiv$ $0$ $($ $mod$ $7$ $)$

Simplifying:

2k \equiv -4 \ (\text{mod} \ 7)

$2$ $k$ $\equiv$ $-4$ $($ $mod$ $7$ $)$ $2k \equiv 3 \ (\text{mod} \ 7)$ $2$ $k$ $\equiv$ $3$ $($ $mod$ $7$ $)$

Next, multiply both sides by the modular inverse of 2 modulo 7. The inverse of 2 modulo 7 is 4 (since $2 \times 4 = 8 \equiv 1 \ (\text{mod} \ 7)$ $2$ $\times$ $4$ $=$ $8$ $\equiv$ $1$ $($ $mod$ $7$ $)$ ). Multiplying both sides of the equation by 4: