# What's the least amount of change I would have to take around with me to always get back less change when I purchase something?

Asked by inunsure (423) September 13th, 2011

Least amount of change as in least coins

Observing members: 0 Composing members: 0

99 cents. Nothing less will do. If something costs 99 cents and you don’t have it, you will have to pay with a dollar and get an extra penny.

If you carry at least 99 cents in change, then for any purchase you will always end up with a dollar less in change than if you did not have enough change to cover the cents in the price. If you start the day with at least 99 cents in change and make multiple purchases, you will always end the day with at most the amount of change you started with.

If you are mainly concerned about accumulating pennies, you can always end up with fewer pennies if you carry 4 of them.

I can’t remember the best solution but I’m sure that isnt it at all, I’m getting there with the answer now

inunsure (423)

\$1.04

Kayak8 (16433)

The entire point of the question is to NOT find yourself having to carry around 99¢ in change. Going out the door with that much defeats the purpose.

Social Engineering:
If something rings up where I’ll get back over 95¢ in change, I make a big deal about patting my pockets and looking for coins. Then I apologize about not having any – implying that it sucks for the cashier to have to count out all the coins needed to make change. More than half the time, the cashier smiles and gives me back a whole dollar.

Check the Environment:
Look for the “Have a penny, take a penny” collection near the cash register. They’re almost always present at gas stations and independent retailers.

Pocket change:
I don’t mind getting back three quarters. What I despise are the pennies and nickels and dimes needed for more than 75¢ return change. So, I’d probably carry around 1 dime, 1 nickel and 3 pennies. If you find yourself wishing you had 4 pennies because 3 isn’t enough, try the Social Engineering bit.

Plastic:
Using credit and debit cards actually increases the cost of goods. There’s always a small surcharge that the retailer must pay for everything purchased with a bank card. That ends up built into the cost of the goods you’re buying. For small mom & pop stores, I try not to use plastic… and they usually have the penny dishes. But at a place like Wal-Mart, I’ll use my debit card. There’s always some cost for convenience.

End of the Day:
If you’re stuck with a lot of change at the end of every day, then make the most of it. When I get home, I put all my change in a huge pickle jar. It fills up pretty fast. And when I cash it out, I usually collect over \$200.

robmandu (21242)

4 pennies will help a lot.

JLeslie (54554)

syz (35523)

Here is another approach for those who do not like carrying much change. Make sure you start out with at least 50 cents. Assuming that all change amounts are equally likely (which may very well not be true), you will on average give up more change than you add.

Or….instead of worrying about how much change you carry, start a change jar.
Mine is only for Quarters, but, I have almost \$20 in Quarters in a little under 2 months of tossing them in a jar when I remember.

Saving dimes and quarters adds up fast.

Coloma (47015)

By the way, this really is not a problem for me because I just charge everything. That can solve your problem, don’t use a lot of cash. \$20 sits in my wallet for weeks sometimes.

JLeslie (54554)

I don’t understand @LostInParadise‘s answer. The whole point is that you use the change you do have to purchase. You don’t just use dollar bills. Let’s say you’re carrying 19¢, in the form of a dime, a nickel, and four pennies. That’s six coins. Something costs 99¢? Okay, so you hand him a dollar, the nickel and the four pennies. You get back a single dime. Now you only have two coins. Thus, you have less change than when you started.

MrItty (17366)

@MrItty

I subscribe to your approach. And maybe I should have ‘interpreted’ what the questioner meant, as you did: “How can I avoid having to carry a lot of change?” because your response correctly addresses that question. But he specifically asked – and most of us still recognize that words have actual meanings, and responded to the request – “to get less change back” during [cash] purchases. I would have interpreted as you did – and as the questioner apparently intended, as he clarified in the thread, but I was feeling persnickety and precise, so I agreed with the answer that @LostInParadise gave, which is the correct response to the literal question.

If the aim is to “get less change back”, then a purchaser has to carry at least the amount of change being requested. Personally, I go with your idea, and try to use the little bit of change that I might have from time to time, add a dollar to the total purchase price – plus my change – and get back “fewer pennies, nickels and dimes” (in that order), which is my normal goal.

CWOTUS (25490)

@MrItty , 1.19 – .99 = .20. That is 20 cents back. You end up with one cent more in change than when you started.

@LostInParadise your math is off. One dollar, the nickel, the four pennies. That’ s \$1.09, not \$1.19.

@CWOTUS If you read the details rather than just the question, you’ll see that the asker clarified “Least amount of change as in least coins”. Unless of course he added that detail after you and LIP posted your answers, in which case, nevermind.

MrItty (17366)

Well, yeah, @MrItty “least amount of change as in less coins” to be received in change is how I read the ‘clarification’ detail. (I’m sure it was present when I wrote my response.) But that’s not how you (probably correctly) interpreted his request to “how can I carry the least amount of change?”, regardless (or irregardless, if you’re minding that thread, too) of “how much change is received”.

CWOTUS (25490)

@MrItty , Okay then \$1.09. You give 9 cents in change and get back 10 cents. You are still up by one cent.

Here is a proof for how this works for those familiar with modular arithmetic.
Let c = cost of item and p = net payment, both expressed in cents

p = c

p = c (mod 100)
The numbers mod 100 are just the cents part of the cost.
In the 99 cent case, we would have
p = 99 (mod 100)
This statement will still be correct if we add or subtract multiples of 100 on the right side.
p = 99 + 100k (mod 100)
All this says so far is that the payment involves some number of dollars + 99 cents

We can write this as
p = 99 – 100 + 100n (mod 100)
p = -1 + 100n (mod 100)

The -1 represents one cent in change.

There is no other way of adding or subtracting multiples of 100 to get a number whose absolute value is less than 100. Therefore the payment involves paying some multiple of dollars and 99 cents or getting change of one cent.

Remember p is net payment. Paying 9 cents and getting back 10 is a net increase in change of one additional cent.

8 coins
.69. 2 quarters, 1 dime, 2 nickle, 4 pennies.

just sayin

blueiiznh (16658)

@blueiiznh I don’t understand two nickels. Two nickels is a dime. Why not one 5 cent piece?

JLeslie (54554)

@LostInParadise Again, you’re not using the asker’s definition of “less change”. You’re interpreting it to mean “the value of the change received, in cents”. Whereas what he meant was “total number of coins, regardless of their monetary value”.

MrItty (17366)

Oh, I misread the problem. I will have to give it some thought.

The best I have come up with so far is nine coins. Half dollar, quarter, dime, nickel and 5 pennies. When I am more awake I might be able to get that down to 4 pennies.

4 pennies (of course)
1 nickel (anything ending in 5 excluding 25, or 75)
2 dimes (complete any ten that a quarter wouldn’t solve)
3 quarters (try it now, you can now make any number under \$1 using 10 coins!)

\$1.04 – many people misunderstood this and thought it was how much makes a dollar or whatev…

ccaron2 (1)

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