What is the smallest number that will satisfy these conditions

Question

I am thinking of a three digit number. If you subtract five from it, the result is divisible by five. If you subtract six from it, The result is divisible by six. If you subtract seven from it, the result is divisible by seven. What is the smallest number that will satisfy these conditions ?

Answer 1

The easiest approach is to understand what being a multiple of a number means.

N-5 is divisible by 5

N-6 is divisible by 6

N-7 is divisible by 7.

 

Meaning 

N-5 is a multiple of 5, so

N-5 = m5

the m means multiple and is an integer.

so n has to be a multiple of 5 too!
Same for the other subtractions:

N-6 is a multiple of 6

N-6 = m6 where m is an integer,

so N is a multiple of 6 too!

N-7 is a multiple of 7

N-7 = m7 where m is an integer,

so N is a multiple of 7 too.

So N has to have factors of 5,6,7 

it's supposed to be the smallest number that will work, so let m = 1

and N = 5*6*7 = 210

 

Here's another thought

 

(N-5) is a multiple of 5, multiples of 5 end in a 0 or 5, so (N-5) ends in a 0 or a 5

So N ends in a 0 or a 5.

Our number is either

ab5

or 

ab0

 

Next 

N-6 is divisible by 6 .
For a number to be divisible by 6 it has to be both even and divisible by three.

So N-6 has to be an even number.
6 is an even number.

 

Recall that 

odd-odd = even

even-even = even

 

n-6=even

means that n has to be even

 

abc has to be even, 

The choice ending in a 0 is even.

So our number is

ab0

 

and you'd continue on,

so that gives you a confirmation of the first presentation, which is the easier path.

 

 

 

 

 

 

 

 

 

 

 

Answer 2

Hi Tina,

 

This is a tricky (round about) way of asking for the LCM (least common multiple of these three numbers.

 

5 * 6 * 7 = 210

 

Here is the wrong way to approach such a simple question

x - 5 = 5p

x - 6 = 6q

x - 7 = 7r

now you have 3 equations and four variables