Here's a great math challenge... is the following number divisible by 3 or not?

Question

123456789101112131415161718192
122232425262728293031323334363
738394041424344454647484950515
253545556575859606162636465666
768697071727374757677787980818
283848586878889909192939495969
79899100


The first person to answer this and explain how he/she did it gets best answer. Yes or No is not enough...

Answer 1

Add the numbers 1 to 100

Sum = n (n+1) / 2

= 100 * 101 / 2

= 5050

Adding 5 + 0 + 5 + 0 = 10 which is not exactly divisible by 3.

So the given number is not EXACTLYdivisible by 3.

Answer 2

No. You can add up all the digits and if the sum is divisible by three, then the number is. For example, 123 can be analyzed by adding 1+2+3 which equals six; six is divisible by 3, thus, so is 123.

I entered the digits of the question cell by cell on an excel spreadsheet and adde them up.

Add up those numbers and you get 890, which is not divisible by three.

=====typing error?====

There is a pattern to the numbers --the digits in the question are all the numbers between 1 and 100 inclusive EXCEPT for 20 and 35. Is this a typing error? Because of you take 20 (2+0) and add 35 (3+5) you get 10 and 10+890 - 900 which is divisible by three and therefore if the numerical sequence in the question was MISTYPED and should have included 20 at the end of the first line and 35 near the end of the second line, then YES that number is divisible by 3 becuase the sum of its digits, 900, is divisible by 3 with no remainder.

Answer 3

Of course, by casting out nines you can simplify the solution. You do not need to sum all the digits. First recognize that the sum of the digits from 1 to 9 is 45 (good old (n+1)*n/2) 45 is divisible by 9 (4+5 = 9) so all of the 1-9 sequences can be tossed out of the digit sum. You have 10 of the 1-9 sequnces in the second digit of each number from 01-09 up to 91-99. There are 10 of each first digits (10, 20, 30....90, 11,21,31...91, etc.) so those can discarded as 10 digit sums as well. This leaves the numeral 1 from the last number (100) so the remainder after division by 3 is 1. It is not divisible by 3 and it does not require a calculator to prove it.

Answer 4

Add all the digits of the given number and see whether the result is divisible by 3. If it is, then the given number is divisible by 3. Otherwise, the given number is not a multiple of 3
Now, I would love to sit and add all these freakin' digits, then divide the result by 3, but I got s h i t to do !!!

Answer 5

yes, because if you add all the digits, you get 1620. Add all the digits again to get nine. Since nine is divisible by 3, then so is the number above. the answer to the problem is 540. that again equals nine. then 180. digits again equal nine. finally you get sixty. The digits equal 6 which is divisible by three. Therefore, it works!!

Answer 6

The answer is no. The way I did it is that I went to the trouble of typing the whole stupid sequence into my computer's calculator program and told it to do division by 3. It did not come out even so 3 is not a factor.

Answer 7

Add up all numbers in number,get total,doagain until number small enough (>100)divide by three, No remainer = Divisible by three.
I just cant be bothered doing it.

Answer 8

Hi. The answer is yes and equals

411522630337040438050539060
640407441417542427643437744
447879127980138081148182158
283168384178485188586198687
208788218889228990239091249
192259293269394279495289596
2996973097983198993299700

I used Karen's Power Tools Calculator program.

The whole answer does not display without a CR!

Answer 9

NO!! Just add up all the numbers and then divide that by three and that's your answer.

Answer 10

yes 1/3 of 12345678910111213141516171819212223242526272829303132333436373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100 is 41152271355594572546113205848795661548795