Consider a collection of pennies with the following constraints: 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 put in groups of seven there are no pennies left over. how many pennies could there be?

91 pennies.

Step-by-step explanation:

It is given that 5 types of groups of pennies: group of 2, group of 3, group of 5, group of 6, and group of 7. If the pennies are arranged in the groups of 7, no pennies are left over. This means that the number of pennies has to be a multiple of 7 in order to satisfy this constraint. In the rest of the groups, there will always be 1 penny remaining without any remaining. This means that the number of pennies will yield the remainder of 1 if it is divided by 2, 3, 5, and 6. Possible multiples of 7:

0, 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98...

The number which satisfies the above conditions is 91. Since 91 divided by 7 is 13 and if 91 is divided by other numbers, the remainder will always be 1. Therefore, there are 91 pennies!!!

3/5

step-by-step explanation:

double number line is not a standard concept, so it must have been defined in your textbook. assume top line is numerator.

the multiples of 15 are 15, 30,45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195,   210, 225, 240, 255, 270, 285, 300, 315, 330, 345, 360, 375,   390, 405, 420, 435, 450, 465, 480, 495, 510, 525, 540, 555, 570, 585, 600, 615, 630, 645, 660, 675, 690, 705, 720, and 735.

the factors of 15 are 1, 3, 5, and 15.

hope that , fam!

-une, your fren : )