This might be very basic but... Take any number for example 2356 now 2+3+5+6=16 and 1+6=7 or 236+5=241 and 2+4+1=7 and also 21+4=25 again 2+5=7 or 652+3=655 which digits sum is 7 again and then 65+5=70 again 7 or 56+5=61 again 7 And so on make any number out of 2356 it will always end up in 7 when digits are added. Similar is true for any other number however big or small. just wondering why?
2026-04-01 14:09:41.1775052581
Basic Maths...can’t get my head around this.
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1
What you appear to be noticing is a result of the following:
Any integer $n$ has the same remainder when divided by nine as the remainder that the sum of the digits has when divided by nine. Similarly so, by repeating the process you get that any integer $n$ has the same remainder when divided by nine as the sum of the sum of the sum of the digits, repeated as many times as is necessary to arrive at a one digit number.
This is an efficient way to check to see if something is a multiple of three, for example $12345654321\mapsto 1+2+3+4+5+6+5+4+3+2+1 = 36\mapsto 9$ so indeed $12345654321$ has a sum of sum of digits equal to $9$ so it is a multiple of nine and therefore also a multiple of $3$.
In your specific observation, you are taking a number, for example $2356$ and adding the digits repeatedly until arriving at a one digit number, in this case $7$. You get the same result whether you added the digits individually, or whether you added them concatenated as in $23+56$ and then repeating the process., this again is because the concatenations of the numbers still have the same remainders when divided by nine as the sum of the digits themselves.