Question

Can anyone please teach me the way of solving this problem? A store runs a promotion in which every 500th customer (500th, 1000th, 1500th, ...) received a free gift, every 800th (800th, 1600th, ...) customer receives a voucher, and every 1200th (1200th, 2400th, ...) customer receives a discount. Which lucky customer will be the first to receive all three promotional prizes?

Answer 1

You must use the Lease Common Multiple technique.

We factor them first...

500 = 5 x 5 x 5 x 2 x 2

800 = 5 x 5 x 2 x 2 x 2 x 2 x 2

1,200 = 2 x 2 x 2 x 2 x 3x 5 x 5

Then we express them with exponents:

500 = 5³ x 2²

800 = 2⁵ x 5²

1,200 = 2⁴ x 5² x 3

Now we multiply the biggest exponents.

5³ x 2⁵ x 3 = 12,000

Answer: The 12,000th customer. Hoped this helped :)

Answer 2

let me put a little differently.

the lucky customer will be one that is divisible by 500, 800 and 1200 simultaneously, namely that is the LCD, Least Common Divisor.

now, we can simplify that a little, let's chop off two zeros from all, and only use 5, 8 and 12, to get the LCD and then we'll put back the two zeros, you can think of getting a common factor of 100, 100(5, 8, 12).

so now let's do some quick prime factoring on them. Let's find the LCD of 5 and 8.

5 = 5 ........... only factor

8 = 2*2*2

5 and 8 have no factors in common, thus their LCD is just 5*2*2*2, 40.

Let's find the LCD of 40 and 12

40 = 5*2*2*2

12 = 2*2*3

the 2 is twice in 12, and twice in 40, therefore, we'll use it in the LCD only once, namely we won't use 2*2*2*2, instead we'll only use 2*2. The 40 has an extra 2 and a 5, we'll use those and the 12 has an extra 3, we'll use that.

so, using the repeated factors in both only once, we end up with an LCD of 2*2*2*5*3 = 120, and if we put back the two zeros, that's 12000.

the 12000th customer is divisible by 500, 800 and 1200 simultaneously, because 12000 ÷ 500 = 24, 12000 ÷ 800 = 15, and 12000 ÷ 1200 = 10.

so she'll get a free gift, a voucher and a discount, and maybe even some confetti.