Find the smallest positive integer n such that the digit sum of n is divisible by 5, and the digit sum of n +1 is also divisible by 5. Note: The digit sum of 1440 is 1+4+4+0 = 9.

Notice that a number is only divisible by 5 if it ends in 5 or 0.

Only using this, we will find that the number is 49,999

Suppose we wanted the sum of the digits to end in 5, then we could have something like:

23

2 + 3 = 5

if we add 1, we get:

23 + 1 = 24

2 + 4 = 6

This will happen always.

instead, if we use a 0 as the last digit of the sum, we get:

For example, the simplest one:

19

we have:

1 + 9 = 10

10 is divisible by 5.

if we add 1, we get:

19 + 1 = 20

2 + 0 = 2

2 is not divisible by 5.

We have a  problem again, but this time you can see that the whole structure of the number does change. So we should keep trying with ones and nines.

Now we need to modify our number to the point we get n + 1 to be divisible by 5.

if we add a 9 to the right side of our number, we get:

199

1 + 9 + 9 = 19

clearly, not divisible by 5.

Here we need to add a 1 at the left side to make it divisible by one, so we get:

1199

1 + 9 + 1 + 9 = 20

this is divisible by 5.

now let's add 1:

1,199 + 1 = 1,200

1 + 2 = 3

3 is not divisible by 5.

Now, if you look at the two numbers we used, we can see that:

for 19, the sum of the digits + 1 gave us a 2.

for 1,199, the sum of the digits +1 gave us a 3.

then just let's add two ones and two nines.

We will get:

11,119,999

adding the digits:

1 + 1 + 1 + 1 + 9 +9 +9 +9 = 40

is divisible by 5.

Adding one to our number, we can get:

11,119,999 + 1= 11,120,000

Adding the digits:

1 + 1 + 1 + 2 + 0 + 0+ 0+ 0 = 5

divisible by 5.

Now, we can group the ones together, so instead of the number 11,119,999

we can have 49,999, which is way smaller.

Adding the digits, we get:

4 + 9 + 9 + 9 + 9 = 40

And adding one to our number, we get:

49,999 + 1 = 50,000

Adding the digits:

5 + 0 + 0 +0 + 0 = 5

So we found the smallest number n such that the sum of the digits of n is divisible by 5, and the sum of the digits of n + 1 is also divisible by 5.

If you want to learn more, you can read:

139,999

Step-by-step explanation:

If the digit sum of n is divisible by 5, the digit sum of n+1 can't physically be divisble by 5, unless we utilise 9's at the end, this way whenever we take a number in the tens (i.e. 19), the n+1 will be 1 off being divisble, so if we take a number in the hundreds, (109, remember it must have as many 9's at the end as possible) the n+1 will be 2 off being divisble, so continuing this into the thousands being three, tenthousands being 4, the hundred thousands will be 5 off (or also divisble by 5). So if we stick a 1 in the beginning (for the lowest value), and fill the last digits with 9's, we by process of elimination realise that the tenthousands digit must be 3 such that the digit sum is divisible by 5, therefore we get 139,999

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step-by-step explanation:

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step-by-step explanation: