Big number multipication

[TimG]

I have a 4th grade student who wants to perform large number multiplication. Earlier in the year, he used a lattice to multiply a 30 digit number by a 30 digit number, now he has even bigger aspirations. He has asked me to provide him with the numbers to multiply together. Specifically, he asked for a 160 digit number and an 80 digit number (apparently how much space he has for a lattice on the large sheet of paper I gave him), but I'd be happy to reduce the size of the numbers by a bit.

 

I'd like to be able to provide him with two numbers that produce a product with a digit pattern of some sort (for one reason it would be easy to check the answer). But, I'm at a bit of a loss as to how to go about coming up with these numbers. Any ideas?

[gwnn]

Wolfram Alpha has infinite precision so checking is not an issue. I can't think of any cool pattern now but anyway if there is some periodicity you are not giving him what he asked for (his task will be simpler than a general one).

 

BTW why can't he find his own 160-digit numbers?

[mattias]

I have a 4th grade student who wants to perform large number multiplication. Earlier in the year, he used a lattice to multiply a 30 digit number by a 30 digit number, now he has even bigger aspirations. He has asked me to provide him with the numbers to multiply together. Specifically, he asked for a 160 digit number and an 80 digit number (apparently how much space he has for a lattice on the large sheet of paper I gave him), but I'd be happy to reduce the size of the numbers by a bit.

 

I'd like to be able to provide him with two numbers that produce a product with a digit pattern of some sort (for one reason it would be easy to check the answer). But, I'm at a bit of a loss as to how to go about coming up with these numbers. Any ideas?

 

Here is a big number calculator that would at least let you check the calculation: http://world.std.com/~reinhold/BigNumCalc.html

[EricK]

174104052846943546074674132671837392418712648437400344812505854633694085663543521

x

638187964577653444444445082632409022097888888888888888888888888888888888888888888888888888888888888888888888888888888888250700924311235444444443806256479866791

 

consists of 240 1's

 

I used the very handy big number factoriser from here http://www.alpertron.com.ar/ECM.HTM

 

(it's worth checking I haven't made a copy/paste error)

[TimG]

Wolfram Alpha has infinite precision so checking is not an issue. I can't think of any cool pattern now but anyway if there is some periodicity you are not giving him what he asked for (his task will be simpler than a general one).

 

BTW why can't he find his own 160-digit numbers?

 

Yeah, I've found big number calculators online and played around a bit (that's also how I checked his 30x30). I'm looking for a neat product.

 

If he gets the numbers from me, he gets a little more attention than if he does it by himself.

 

I gave him 300 digits (from pi) that he can split up how he wants. But, there is time to give him a new pair of numbers tomorrow.

[gwnn]

For example, suppose you use mattias' product. Do you think your student will appreciate it when he sees that the first (or last) 20-30 digits are all 1's? It seems that it makes his job much simpler than what he asked for (he can check any digit instantly). Of course you can use a more complicated pattern but it sounds like the dude is quite smart so I would expect he crack the neater ones easily.

[EricK]

For example, suppose you use mattias' product. Do you think your student will appreciate it when he sees that the first (or last) 20-30 digits are all 1's? It seems that it makes his job much simpler than what he asked for (he can check any digit instantly). Of course you can use a more complicated pattern but it sounds like the dude is quite smart so I would expect he crack the neater ones easily.

But if he wants to actually do the multiplication rather than just get the answer (and judging by the OP he does) then what does it benefit him to cheat?

 

Does the student realise how long the multiplication of a 160 digit number by a 80 digit number will take?!

[gwnn]

But if he wants to actually do the multiplication rather than just get the answer (and judging by the OP he does) then what does it benefit him to cheat?

 

Does the student realise how long the multiplication of a 160 digit number by a 80 digit number will take?!

I was not talking about cheating. I was talking about him knowing what the next digit will be so if he makes a mistake he will immediately find it, whether he wants to or not. And if there is a periodicity, it may mean that some (many) of the sums will be identical or similar, so it will not be as interesting for the student as it would be with two random numbers.

 

When I asked 'will your student appreciate it if (he notices the pattern)?' I meant to suggest that he would NOT appreciate it. Sorry for the confusion caused by my wannabe rhetoric.

[gwnn]

Oops and I mixed up two repliers. Sorry about that too.

[mgoetze]

it sounds like the dude is quite smart

I dunno... I expect the truly smart ones to figure out that computers are better at this chore sooner or later. :P

[BunnyGo]

You could give him:

 

14894830945427842136024287345244264001934112429134263004678087064894905384501

 

times

 

67137385020604260224163267966933051652653639791893037880343374935793363268430985878726149793957397758367320330669483473463602081069621196566250642066367315690134499

 

Should be 10^240 - 1 so just 239 "9"s

[JLOGIC]

Best thread ever

[kenberg]

Perhaps

 

225319534991831177328890236228992001350685163362356544091910

 

times

 

1490866825405028349797207283090793766362403207759454948507817712458718\

5935174056446232828346754216690615964568306082260925528298256098234372\

15523801709217726693

 

 

The first number is the product of the first 35 primes: 2X3X5X...X151

 

The second number is the product of the primes from the 40th prime through the 102 nd prime: 173X...X557

 

 

Or so I think.

 

He may as well learn a little about primes while doing all of this multiplication. Here are the first 102 of them. He may be amused to find the twin primes (primes differing by 2).

 

 

{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61,

67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137,

139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211,

223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283,

293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379,

383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461,

463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557}

 

 

The product of the two numbers is

3359214198350085477151500518185826896761297540250716248287028945429425\

1747607029264887200132433081899785613575258298534495659537412581004401\

6223369623555270410256256504137112460740389892294939832396376962106230\

252353630

[Mbodell]

I think perhaps it is time to move him on to other questions with him like:

 

How many different 160-digit by 80-digit multiplication questions could you give him?

In doing the multiplication how many single-digit by single-digit operations would he do (assuming that is how he's doing it)?

How many different single digit by single digit could there be, counting carries (so if there is some place he needs to take a 4 and multiply 7 with no carry, that is one. A 4 multiplied by 7 with a carry of 1 would be another. A 4 multiplied by 7 with no carry would be a repeat and wouldn't count additionally)?

Is it possible for him to construct 2 different 160-digit and 80-digit multiplications that both give the same answer? 3 different with the same answer? 4 different with the same answer? More? Is there anything that they have in common?

Can he prove that 1234567890 repeated 23 times is not a possible answer to any 160-digit by 80-digit multiplication?

Can he prove that 9876543210 repeated 30 times is not a possible answer to any 160-digit by 80-digit multiplication?

Can he estimate how many possible numbers which are answers to a 160-digit by 80-digit multiplication?

Can he prove that 170141183460469231731687303715884105727 is not a possible answer to any 16-digit number multiplied by a 23-digit number?

Does that change his estimate to how many possible numbers which are answers to a 160-digit by 80-digit multiplication?

Can he figure out how many different rearrangements there are of the letters of his name? The name of his state?

[kenberg]

I would like to hear how this comes out. "Eyes bigger than his stomach" comes to mind, but perhaps he will actually complete the calculation. Me, I collected pet frogs. But not 60 of them.

[TimG]

I would like to hear how this comes out. "Eyes bigger than his stomach" comes to mind, but perhaps he will actually complete the calculation. Me, I collected pet frogs. But not 60 of them.

 

Earlier in the year, he completed a 30 digit x 30 digit multiplication problem. He is using the lattice method, so he did 900 single digit x single digit multiplications and then added 60 columns (diagonals) of one digit numbers. No, he did not get it all right the first time through. He and I checked all 900 products together. I told him which column sums were incorrect.

 

This is not a gifted student (if he was he'd probably be doing some of the things Mbodell suggested) it's more an OCD thing.

[TimG]

That's just the sort of thing I was looking for. Thanks.

 

You could give him:

 

14894830945427842136024287345244264001934112429134263004678087064894905384501

 

times

 

67137385020604260224163267966933051652653639791893037880343374935793363268430985878726149793957397758367320330669483473463602081069621196566250642066367315690134499

 

Should be 10^240 - 1 so just 239 "9"s

[ArtK78]

A friend of mine at Princeton was able to do arithmetic calculations essentially instantaneously. And I don't mean 4+2. I would ask him to give me the 26th root of 2692752156 (and similarly bizarre computations) and he would immediately provide me with the correct answer (confirmed by my pocket calculator).

 

I have no idea how he did that. It must have been some inate ability.

 

He learned bridge from me essentially from scratch. He became competent in very short order, but I don't think he ever pursued it beyond the club level.

 

He was no "idiot savant." After graduating Princeton (at the age of 18) with a degree in mathematics, he went on to Harvard Law School (Law Review) and is a partner in a prominent law firm in Washington, DC. I occasionally hear about him when he argues a case before the Supreme Court.