the product of two consecutive negative integers is 272. find the larger of the two integers.

if 3/x+2 = y+3 /2 and x and y are integers, what is the sum of all possible values x?

A chicago subway card is worth $11. For each ride on the subway, either $1.50 or $1.80 is deducted from the value remaining on the card. what is the least number of cents that could be left on the card after any number of rides?

hector spent a total of exactly $200 on shirts and pants. The shirts cost $!5 each and the pants cost $22 each. how many shirts did hector buy?

If it is now 6 AM Tuesday, what day of the week will it be in 50,000 minutes?

Guest Sep 15, 2022

#1**0 **

*the product of two consecutive negative integers is 272. find the larger of the two integers.*

Ignoring the polarity for the moment.

Since they're consecutive, they're close to each other, but more important,

they're close to the square root. So what's the square root of 272?

If you don't know, you can use the calculator in your phone to find it. It's about 16**.**49.

So what two consecutive numbers close to 16 multiply to 272?

Note the final digits of those two numbers will have to multiply to a number that ends in 2.

How about 6 and 7, that's 42 so it sounds like a winner. So are the numbers 16 and 17?

Does 16 x 17 = 272? Try it and see. **YES**.

Now going back, we have to remember that the polarity is negative.

So, the larger number is the one that's less negative, so our answer is **–16**.

_{.}

Guest Sep 16, 2022

#2**0 **

*A chicago subway card is worth $11. For each ride on the subway, either $1.50 or $1.80 is deducted from the value remaining on the card. what is the least number of cents that could be left on the card after any number of rides?*

Assume you spend 1**.**50 as many times as you can.

You ride 7 times for a total of 10**.**50 so you have 0**.**50 left over.

You can use up another 0**.**30 of that 0**.**50 by substituting one 1**.**80 ride.

Six rides at 1**.**50 plus one ride at 1**.**80 costs you 9**.**00 + 1**.**80 = 10**.**80. 20¢ left on the card.

Let's put the totals in a tabulation, using a decreasing number of 1**.**50 rides.

(7)(1**.**50) + (0)(1**.**80) = 10**.**50 + 0 = 10**.**50 0**.**50 left

(6)(1**.**50) + (1)(1**.**80) = 9**.**00 + 1**.**80 = 10**.**80 0**.**20 left

(5)(1**.**50) + (1)(1**.**80) = 7**.**50 + 1**.**80 = 9**.**30 1**.**70 left

(4)(1**.**50) + (2)(1**.**80) = 6**.**00 + 3**.**60 = 9**.**60 1**.**40 left

(3)(1**.**50) + (3)(1**.**80) = 4**.**50 + 5**.**40 = 9**.**90 1**.**10 left

(2)(1**.**50) + (4)(1**.**80) = 3**.**00 + 7**.**20 = 10**.**20 0**.**80 left

(1)(1**.**50) + (5)(1**.**80) = 1**.**50 + 9**.**00 = 10**.**50 0**.**50 left

(0)(1**.**50) + (6)(1**.**80) = 0 + 10**.**80 = 10**.**80 0**.**20 left

The table also contains an increasing number of 1**.**80 rides, so

all combinations of rides totaling nearly 11**.**00 are accounted for.

So the least amount that can be left on the card is determined to be **20¢**.

_{.}

Guest Sep 17, 2022

#3**0 **

*hector spent a total of exactly $200 on shirts and pants. The shirts cost $!5 each and the pants cost $22 each. how many shirts did hector buy?*

Look at the total, it's 200. It ends with a zero.

So added together, the total of the costs of the shirts and the pants has to end with a zero.

The only digit you can get at the end of the shirts total is a zero or a five.

Five is no good because you can't get the other five as a multiple of 22.

So, the number of shirts times 15 has to end with a zero.

And the number of pants times 22 has to end with a zero, also.

For the pants total to end with zero, it will have to be multiplied by either a 5 or a 0.

Ten is no good because the total would be more than 200 so it has to be 5.

That means Hector bought 5 pants for a total of 110. That leaves 90 for the shirts.

Shirts = 90/15 = **6 shirts**.

_{.}

Guest Sep 17, 2022