Afive-digit number is represented by abcde. if we add the number 1 in front of abcde, then the product of 1abcde and 3 will be the six-digit number abcde1. what is the original five-digit number abcde?

ABCDE = 42857

Step-by-step explanation:

First, we will use logic. The only number that when it's multiplied by 3, gives 1 at the end is 7, so E should be 7, so:

1ABCD7 * 3 = ABCD71    

If 3 times 7 is 21, we carry two, so, the next number by logic, cannot be 1, because 3*1 = 3 + 2 = 5, it's not 7, so, it should be another number, like 5.

5*3 = 15 + 2 = 17 and we carry one. So this number fix in the digit, and D = 5.

We have now: 1ABC57 * 3 = ABC571

Letter C, we have to get a number that when it's multiplied by 3 and carry one, gives 5. In this case 8, because: 3 * 8 = 24 + 1 = 25 and carry two. D = 8.

So far: 1AB857 * 3 = AB8571

Now, the same thing with B. If we multiply 3 by 2, and carry two we will have 8 so: 3 * 2 = 6 + 2 = 8. B = 2

1A2857 * 3 = A28571

Finally for the last number, a number multiplied by 3 that hold the 1 as decene, In this case, the only possibility is 4, 3 * 4 = 12 so:

142857 * 3 = 428571      

hello from mrbilldoesmath!

answer:   x (x+6)^2, choice b

discussion:

factor x out of the equation:

x^3 + 12x^2 + 36x = x (x^2 + 12x + 36).

note that 6+6 = 12 and 6*6 = 36 so the above factors into:

x (x+6)^2

which is choice b

you,

mrb