Aida's remainder

From Euclid's division $\frac{a}{b} = q + r \Rightarrow$ $0 \le r < a$

Maximum value = $a + b$ can have is $18$ . So $r \le 17$ .

$a + b = 18 \rightarrow a = 9 \to b = 9$

$99$ mod $18 = 9$ .

$a + b = 17 \rightarrow a = 9,8 \to b = 8,9$

$98$ mod $17 = 13$

$89$ mod $17 = 4$

$a + b = 16 \rightarrow a = 9,8,7 \to b = 7,8,9$

$97$ mod $16 = 1$

$88$ mod $16 = 8$

$79$ mod $16 = 15$ .

If we carry on with more examples we'd never find any remainder equal or higher than 15 because we know $r < a + b$ , and if we continue $a + b = 15$

The largest possible digit sum of digits = 9 + 9 = 18,

so no remainder can be less than 17.

Working down from the top:

99/18 → remainder = 9

98/17 → remainder = 13

89/17 → remainder = 4

97/16 → remainder = 1

88/16 → remainder = 8

79/16 → remainder = 15

After that the divisor is 15 or less,

so the remainder is 14 or less,

and so the largest remainder is 15,

from 79 = 4 * 16 + 15.

To get the largest possible remainder we need to start with numbers whose digits add up to a high number. The highest possible such number is 99, whose digits add up to 18.

$99 ÷ 18 = 5 R 9$

89 and 98's digits add to 17:

$89 ÷ 17 = 5 R 4$

$98 ÷ 17 = 5 R 13$

13 might seem high enough to you, but we can't be so sure! 79, 88, and 98's digits add to 16:

$79 ÷ 16 = 5 R 15$

$88 ÷ 16 = 5 R 8$

$97 ÷ 16 = 5 R 1$

Notice how in $79 ÷ 16$ , the remainder is one less than the divisor. When this is the case, we know we can't go any higher. Therefore, the highest possible remainder is $15$ .

When 79 is divided by the sum of its digits (7+9=16), we get a remainder of 15.

Since 15 is the biggest possible remainder when the sum of the digits is 16, the only way we could get a remainder bigger than 15 is if the sum of the digits were 17 or more.

There are only 3 two digit numbers whose sum of digits is 17 or more and they are 89, 98 and 99. If we examine these 3 cases we will find that the remainder in each case is less than 15.

Therefore the largest possible remainder is 15.

The largest possible digit sum = 9 + 9 = 18, so no remainder can be > 17.

Working down from the top: 99/18 → remainder 9 98/17 → remainder 13 89/17 → remainder 4 97/16 → remainder 1 88/16 → remainder 8 79/16 → remainder 15

After that the divisor is 15 or less, so the remainder is 14 or less, and so the largest remainder is 15

79/16= 4 remainder 15....

To get the required answer we must be clear that--- 1. The 2-digit number is not divisible by the sum of digits So let us consider only the 2 digit primes- Starting from 11, 11 = ( 2 * 5 + 1 ) _ _ _ _ remainder = 1 13 = ( 4 * 3 + 1 ) _ _ _ _ remainder = 1 so on...............................we get the required prime number 79 which bears the largest remainder - 79 = 16 * 4 + 15 so,our answer is 15

N=10 x +y We have to maximize N%(x+y) Which is equal to 9 x%(x+y) If we put x =1 then x+y will vary from 1 to 10 For x= 1 max remainder we get is 9 And for x =2 we have to check only for x+y >=10 because for x+y<10 remainder will be less than 9 . So on we go for x=1 to 9 We get a maximum remainder

1. let's define $AB$ a 2 digit number, $a$ the ten part, $b$ the unit part, $maxR$ the biggest remainder and $minR$ the smallest one
2. we got $AB=k*(a+b)+maxR$ but this is equal to $AB=(k+1)*(a+b)-minR$ with $minR+maxR=a+b$ and $minR=1$ in the best case
3. $a+b=\frac {10*(a+b)}{10}=\frac {10*a+10*b}{10}=\frac{AB+9*b}{10}$ because $AB=10*a+b$ thus $\frac{\frac{AB+9*b}{10}}{a+b}=1\Rightarrow\frac{AB}{a+b}=10-\frac{9*b}{a+b}$
4. the right part of $\frac{AB}{a+b}=10-\frac{9*b}{a+b}$ contains an integer part subtracted by a fractional part that mean that the $minR=1$ is contained in the fractional part so solving the problem is the same thing than solving the equation $9*b=m*(a+b)+1$ for the biggest $a+b$ sum which is much simplier
5. let's try with $b=9$ this gives $80=m*(a+9)$ . $a=7$ gives us $16$ the biggest natural divisor for $b=9$ with $m=5$ thus we should have $maxR=16-1=15$
6. suppose that we have $maxR>15$ this can only happen for $b<9$ and $a+b>17$ in the same time. Impossible because this gives $a>9$
7. so the result is $15$ for $79=4*16+15$

SO here is it..without using brute force...it struck me naturally...the solution...it's lengthy but actually it's not..it reduces possibilites at every step and we approach to the final solution smoothly...criticism and appreciation are welcome.

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The sum of a two digit number's digits can be no more than 18. So the greatest possible remainder is 17?

Unfortunately in reality the answer is 15. Don't ask me how it happens, but here's the practical:

http://dev.byteofk.com/2digit.php

The PHP code to output this if anyone is interested:

<? $largest = 0; for($x=10;$x<100;$x++){ $tens = substr($x,0,1); $ones = substr($x,1,1); $sum = $tens+$ones; $res = floor($x/$sum); $rem = $x-($res*$sum); $display = "$x: $tens + $ones = $sum, $x / $sum = $res rem $rem<br/>"; echo $display; if($rem>$largest){ $largest = $rem; $largestdisplay = $display; } } echo "Largest remainder: $largest<br/>$largestdisplay"; ?>

Let the number equal $a$ and $S(a)$ denote the sum of its digits.

$\implies a = S(a)q + r$ , where $q$ is the quotient, $r$ the remainder, such that $r < S(a)$ .

Therefore, for $r$ to be maximum, $S(a)$ has to be maximum. We test out the largest possible values of $S(a)$ :

$99 = 18\times 5 + 9$

$98 = 17\times 5 + 11$

$89 = 17\times 5 + 4$

$79 = 16\times 4 + \boxed{15}$

Observe that $15$ is the largest value possible, $16$ is the largest digit sum we have encountered so far, and all other digit sums will be less. It follows that all other remainders will be lesser as well.

We can write the problem as $10a+b=q'(a+b)+r$ where $r \in [0,a+b)$ . Setting $q'=q+1$ , we have $9a=q(a+b)+r$ . Basically, $b$ can take any value $[0,9]$ for any value of $a$ . So let $a+b = 18$ . Then $max(rem(9a,18))=9$ . $max(rem(9a,17))=13$ . $max(rem(9a,16))=15$ . Since $max(rem(9a,y)) < 15$ for all $y \in [0,15]$ , 15 is the largest possible remainder.