When the Pennies are put in groups of 2 there is one penny left over.
When they are put in groups of three, five and six there is also one penny left over.
But when they are upt in groups of seven there are no pennies left over.

I need a picture too of an explanation/ example

The number of pennies is 91 is an example for the situation

Step-by-step explanation:

you are looking for a number divisible by 7 and gives reminder 1 when divided by 2, 3, 5 and 6

So let us search in the multiple of 7

Multiples of 7 are 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84 , 91, ...

∵ All even numbers divisible by 2

∴ The multiple of 7 must be odd

∵ The multiple of 7 gives remainder 1 with 2, 3, 5, 6

- Find the common multiple of 2, 3, 5, 6 and add 1 to it

∵ The common multiplies of 2, 3, 5, 6 are 30, 60, 90, 120, 150,

∵ 30 is the 1st common multiple of them

∵ 30 + 1 = 31

∵ 31 is not a multiple of 7

∴ 31 is not the answer

∵ 60 is the 2nd common multiple of them

∵ 60 + 1 = 61

∵ 61 is not a multiple of 7

∴ 61 is not the answer

∵ 90 is the 3rd common multiple of them

∵ 90 + 1 = 91

∵ 91 is a multiple of 7

∴ 91 is divisible by 7 and gives remainder 1 with 2, 3, 5, 6

∵ 91 ÷ 2 = 45 and remainder 1

∵ 91 ÷ 3 = 30 and remainder 1

∵ 91 ÷ 5 = 18 and remainder 1

∵ 91 ÷ 6 = 15 and remainder 1

∵ 91 ÷ 7 = 13 without remainder

∴ The number of pennies is 91

Using multiples and remainder of divisions, it is found that there were 91 coins.

Shortening the possibilities, we look at odd multiples of 7, which are: {7, 21, 35, 49, 63, 77, 91, 105, 119, ...}

A similar problem is given at link

46.2+112.3

step-by-step explanation:

add an !

step-by-step explanation: