Question

5 divides 210M the what is the least value of m​

Answer 1

Answer:

Mathematics

x+2 must be divisible by 3,4,5, and 6. Hence x+2=n⋅lcm(3,4,5,6)=n⋅60, for any integer n. So x=60n−2 for any integer n. Assuming the problem asks for the "smallest positive" such number, the answer is 58.

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answered

Apr 21 '14 at 17:11

Marcus Neal

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Apr 27 '14 at 8:36

Martin Thoma

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Up vote

2

Down vote

So we need x=3a+1=4b+2=5c+3=6d+4

which can also the written as x=3(a+1)−2=4(b+1)−2=5(c+1)−2=6(d+1)−2

So, we need to find x such the remainder =−2 for the divisors 3,4,5,6

Now, the smallest number which is divisible by 3,4,5,6 is lcm(3,4,5,6)=60

So, 60m−2 (where m is an integer) will leave −2 as remainder

Find proper m for the minimum positive value of x

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answered

Mar 29 '14 at 13:39

lab bhattacharjee

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Up vote

1

:

x = 0

while True:

if x % 3 == 1 and x % 4 == 2 and x % 5 == 3 and x % 6 == 4:

print(x)

break

x += 1

A faster solution

Every single of the six constraints has to be true. The sixth constraint is only true for every sixth number, so we can "jump" in steps of six:

x = 4

while True:

if x % 3 == 1 and x % 4 == 2 and x % 5 == 3:

print(x)

break

x += 6

Answer

The answer is 58.

.

Answer 2

Answer:

x=2

y=3

z=4

as per laborately can you tell us about the same time