Question

What is the least number which when divided by the number 3 5 6 8 10 and 12 leaves in each case remainder 2 .but which when dividedby thirteeen leaves no remainder . ?

Answer 1

Question :-

• find the Least Number which when divided by the number 3 5 6 8 10 & 12 leaves in each case remainder 2 but which when divided by 13 leaves no remainder . ?

Solution :-

The Number will be in the form of :- [{(LCM of Numbers) * k + 2 } / 13 ] . where k is any constant Natural Number.

LCM of 3, 5, 6, 8, 10 , 12 :-

→ 3 = 1 * 3

→ 5 = 1 * 5

→ 6 = 2 * 3

→ 8 = 2 * 2 * 2 = 2³

→ 10 = 2 * 5

→ 12 = 2 * 2 * 3 = 2² * 3

LCM = 3 * 5 * 2³ = 120 .

So,

→ The Number That Give Remainder two would be = LCM + 2 = 120 + 2 = 122. But Since it is Not Divisible by 13 .

Therefore, Let Check Next now, or we can put value of k = 1,2,3______

→ [{(LCM of Numbers) * k + 2 } / 13 ]

→ [ (120 * k + 2 ) /13)

Putting k = 2 now,

→ [ (120 * 2 + 2) / 13)

→ ( 240 + 2) / 13

→ 242/13 ≠ Divisible .

Putting k = 3 now,

→ [ (120 * 3 + 2) / 13)

→ ( 360 + 2) / 13

→ 363/13 ≠ Divisible .

Putting k = 4 now,

→ [ (120 * 4 + 2) / 13)

→ ( 480 + 2) / 13

→ 482/13 ≠ Divisible .

Putting k = 5 now,

→ [ (120 * 5 + 2) / 13)

→ ( 600 + 2) / 13

→ 602/13 ≠ Divisible .

Putting k = 6 now,

→ [ (120 * 6 + 2) / 13)

→ ( 720 + 2) / 13

→ 722/13 ≠ Divisible .

Putting k = 7 now,

→ [ (120 * 7 + 2) / 13)

→ ( 849 + 2) / 13

→ 842/13 ≠ Divisible .

Putting k = 8 now,

→ [ (120 * 8 + 2) / 13)

→ ( 960 + 2) / 13

→ 962/13 = Divisible by 13. (Ans.)

Hence, Our Least Required Number will be 962.

(Nice Question.)

Answer 2

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$\huge\tt{TO~FIND:}$

• What is the least number which when divided by the number 3 5 6 8 10 and 12 leaves in each case remainder 2 but which when dividedby thirteeen leaves no remainder . ?

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$\huge\tt{SOLUTION:}$

The number would be like -

[{(LCM of Numbers)×k+2}/13] we know where k is an constant number

Now, LCM = 3×5×2³ = 120

The number which give 2 reminder would be (120+2)=122

but we know that 122 isn't completely divisible by 13

[{(LCM of Numbers)×k+2}/13]

[{120×k+2)/13}]

Let's put the value of k = 8,

[(120×8+2)/13]

[960+2/13]

(926/13)

So, 962 is divisible by 13

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