What digit, in place of the □ , will make the 8-digit integer, 2 3 4 5 6 7 8 □ a multiple of 6?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
I came at the answer somewhat differently. Knowing that the answer has to be either 0, 2, 4, 6, or 8, which are the only final digits possible for an integer multiple of 6, it was also a reasonable assumption due to the wording of the question that there was only ONE answer.
This rules out 0 and 6 since you could just add 6 to get the second answer. Similarly, it also rules out 2 and 8 for the same reason.
This leaves only 4 as a unique answer, since adding or subtracting 6 would change the second last digit as well.
4 is impossible. How come? The answr is 6. Just put 6 at the end of it and divide it by 6. Then you get 39xxxxxxx. However, if you put 4 and divide it by 6, then you get 39xxxx.xyz
If you multiply a number by a number, then the other way aroubd must work and in this case it fails. Example: 1539 × 3 = 4617 >>>> 4617 / 3= 1539 234567874 / 6 = 39094645.666 234567876 / 6 = 39094646. =>> this is the number that you will get if yoy put 6 at the end of that big number.
Log in to reply
the big number is 2 3 4 5 6 7 8 □ , not 2 3 4 5 6 7 8 7 □
35+4=39 39 is not divisible 2 therefore, 7 should be the right answer
Log in to reply
The addition rule is for 3 only not 2
The sum dosen't have to be divisible by 2.. Only the final no
Also, 39 is not divisible by 7 either
to be divisible by 2 it must end in an even number. to be divisible by 3 all of the digits must add up to a number divisible by 3, like 39.
LOL
This addition rule only require to 3,9 or 11 but divisible by 2 is the last number of the number is 0 , 2 , 4 , 6 , 8
the last two digits placed together to make a two digit number in this case: 82 or 84 or 86 or 88 must be a multiple of 6. So for the large number of 2345678# the number must have 84 as the last two digits since 84/6 is=14 (6x10=60 and 6x4=24; 60+24=84)
Thats how i got my answer as well
If you use https://en.wikipedia.org/wiki/Casting out nines you will add digits to use only 1-digit numbers, to check which is multiple of 3. Only the multiple of 3 can be multiple of 6.
I shortened the link just for you http://tiny.cc/tqr7vy
Let 2345678 □ = 23456780 + x
6 - (23456780 modulo 6) = x
6 - 2 = x
4 = x
Hi, Could please explain this modulo rule: "6 - (23456780 modulo 6) = x" and how could you think this particular way. Regards
numbers come in patterns.ignore everything but 8[]. Find n * 6=~80. 14 * 6=84, therefore 4. i had a longer answer but why complicate things.
By divisibility rule.. Let take prime factor 6 that is 2 and 3 if the number is divided by these 2 number it's also divisible by 6.
Now just check divisible rule of 2 : last digit is even number so it's 4 divided by 2 hence it is divisible by 2 now
Divisible rule of 3 (sum of all digits divisible by 3) 2+3+4+5+6+7+8+4 = 39 (divisible by 3)
Hence the number is also divisible by 6 so 4 is our answer.
Taking 8 divisable by 6 we get 2 as remainder,now 24 is divisible by 6 so answer is 4
2345678?
2345678÷6=390946, r=2
since 24 is divisible by 6 therefore the last digit is 4
So the number is 23456784
if a number is divisible by 6,it is even and divisible by 3. here 2+3+4+5+6+7+8=35. thus in this situation we need to find a number divisible by 3 when it has 35 taken away it is even and also is near to 35. In this case it is 39,then 39-35=4 hence the answer is 4.
If 2 3 4 5 6 7 8 □ divisible by 6 then it can divisible by 2:which is 0 , 2 , 4 , 6 , 8 & dividible by 3 which is 3 5 + □ .The only solution is 4
The number is divisible by 6 if it is divisible by both 2 and 3. Thu sum of digits is: 2+3+4+5+6+7+8=35. The sum of all digits must be divisible by 3 and only 35+4=39 is divisible by 3 ( 35+8=43, 35+6=41, 35+2=37 ).
For a number to be divisible by 6, it must be divisible by both 2 and 3. We add the digits up to the mystery digit and get 35. So, we have to add 4 (the mystery digit) so that the number is divisible by both 2 and 3.
The number must be even. If the solution was 0, there would be another solution: 6. If the solution was 2, there would be another solution: 8. Since there is always a unique solution to these questions, the answer is 4. You don't need to sum the digits.
I just divided 2345678 by 6 had a reminder of 2 and knew 24 is the only # divisible by 6 so answer is 4. Why over complicate things.
O número final é 8. Diminuindo 6 resta 2. O número quatro no final é o único que dá o número 24 que é divisível por 6
All you have to is choose an answer that will make the final two digits divisible by 6. Thus making the whole number divisible by 6. So out of the choices the only possible answer is 4.
So you're saying that 784 is divisible by 6? Along with 1084, 584, 5384, etc?
Log in to reply
It might be that he noticed that the first six digits add up to 27, which is divisible by three, so the first six digits (234567) followed by 00 are a multiple of three, multiplied by 100. 100 has a 2 in it, therefore 23456700 is a multiple of six. Then all you have to worry about is that the final two digits are a multiple of six, because if you add two multiples of six you get a multiple of six.
For a number to be divisible by 2, the number must be even, to be divisible by 3 the sum of the digits must be divisible by 3...therefore, if you add the numbers together, you get 35, thus leaving 4 as the right answer
We know that if the last two digits of the number are divisible by 6 then then it is a multiple of 6. Since 84 in 80s is the only divisible by 6, so it's divisible by 6.
this is not true always: 184, 284 and so on
Log in to reply
NOT TRUE. 118 is not divisible by 6 whilst 18 is.
See my response above. I'm not sure if these people are actually not understanding it or if they are just leaving our part of their reasoning. If you add the digits from left to right, you get to 27 at the 6th digit, so you know that 23456700 must be divisible by six. Then you only need to have the last two also divisible by six. I am guessing this is why two people came up with this, but I might be giving them too much credit. There is a correct rule that says the whole number is divisible by 4 if the last two digits are divisible by 4. But that is just a consequence of the fact that XXX00 is a multiple of 4 because 100 is a multiple of 4, that certainly doesn't hold for 6 (or any other number except products of at most two 2's and two 5's) as you have demonstrated.
Problem Loading...
Note Loading...
Set Loading...
If a number is a a multiple of 6, then it is a multiple of 2 and 3.
If 2 3 4 5 6 7 8 □ is a multiple of 2, then the last digit is even which means that □ must be either 0, 2, 4, 6 or 8.
If 2 3 4 5 6 7 8 □ is a multiple of 3, then the sum of the digits is a multiple of 3, or that 3 5 + □ is a multiple of 3. Thus, □ must be either 1, 4 or 7.
The only number that appears in both cases is 4. Thus □ = 4 .