Divisors of a0b

For $b = 0$ , the condition that $\overline{a0}$ divides $\overline{a00}$ holds true for all possible values of $a$ , i.e. $1, 2, ... 9$ . That yields the following solutions for $N$ : $10, 20, ... 90$ .

If $b \neq 0$ , we observe that $\overline{a0b} < \overline{ab0} = 10 * \overline{ab}$ . So, the result of dividing $\overline{a0b}$ by $N$ must be strictly lower than $10$ . We can assign the quotient values from $9$ to $1$ to find additional solutions:

• $\overline{a0b} = 9 * \overline{ab} \Rightarrow 5 * a = 4 * b \Rightarrow N = 45$ is a solution.
  • $\overline{a0b} = 8 * \overline{ab} \Rightarrow 20 * a = 7 * b \Rightarrow$ No solution found.
  • $\overline{a0b} = 7 * \overline{ab} \Rightarrow 5 * a = b \Rightarrow N = 15$ is a solution.
  • $\overline{a0b} = 6 * \overline{ab} \Rightarrow 8 * a = b \Rightarrow N = 18$ is a solution.
  • $\overline{a0b} = 5 * \overline{ab} \Rightarrow 50 * a = 4 * b \Rightarrow$ No solution found.

We could go over the next possibilities ${4, 3, 2, 1}$ , but we can observe instead that the ratio $\frac {b}{a}$ increases as the quotient gets smaller. For a quotient value of $5$ , the ratio already exceeds $10$ . Taking into consideration that $a$ and $b$ are both digits, we can conclude that we won't find any other solutions.

Finally, $(10 + 20 + ... + 90) + 45 + 15 + 18 = 528$ .

Let $n(10a+b)=100a+b$ for some positive integer $n$ . This implies that $a(100-10n)=b(n-1)$ where $1<=n<=10$ . By checking all cases for $n=1,2,3, \ldots, 10$ : $n=1,2,3,4,5,8$ have no solutions.

When $n=6$ , $40a=5b$ , so $8a=b$ , has solution $a=, b=8$ .

When $n=7$ , $30a = 6b$ has solution $a=, b=5$ .

When $n=9$ , $10a=8b$ has solution $a=4, b=5$ .

When $n=10$ , $9b=0$ , has solution $a=1,2,3,...,10$ as long as $b=0$ .

Adding all possible values, $18+15+45+10+20+30+40+50+60+70+80+90=528$

100a+b=k(10a+b) 10(10-k)a=(k-1)b When b=0, then k=10, a=1,2,…,9 When b<>0, then consider the value of k: K=9, 10a=8b, so ab=45 K=8, 20a=7b K=7,30a=6b,so ab=15 K=6,40a=5b,so ab=18 K=5,50a=4b, no solution Sum=10+20+…+90+45+15+18=528

The conditions of the problem are satisfied if and only if there exists an integer k such that k = a 0 b / ab , or k x ab = a 0 b .

Condition is trivially satisfied whenever b =0 (e.g., 200 divides by 20, the case of a =2 and b =0). Summing these nine cases, where a =1,...,9, yields 10 n ( n +1)/2 = 10 x 45 = 450 .

For the remaining cases, namely when b > 0, it can easily be demonstrated that the product k x ab = a 0 b will not maintain the units digit, b , when k is 2, 4, or 8, since multiplication by 2, 4, and 8 always result in a different units digit.

In fact, the only cases where the units digit will be maintained are in the following ( k , b , k x b ) triplets: (9, 5, 5), (7, 5, 5), (6, 4, 4), (6, 6, 6), (6, 8, 8), (5, 5, 5), (3, 5, 5), (1, ALL, ALL)

Let's look more closely at the remaining cases, where b > 0.

We can see that when a = 9, k falls between 901/91 and 909/99. Thus 10 > k > 9, so there are no integer quotients for any of these cases.

Similar for a = 8.

When a is 4, 5, 6, or 7, we can easily see that 10 > k > 8, leaving only 9 as a potential integer quotient. Furthermore, the triplets above would then require that b = 5 when k = 9. Thus only (45, 405), (55, 505), (65, 605), and (75, 705) need to be tested and, in fact, only 405/ 45 yields an integer k .

When a is 2 or 3, we see that 10 > k > 7, leaving only 9 and 8 as potential integer quotients. Recall (or examine the possible triplets to see), however, that multiplication by 8 will never maintain the units digit, so this possibility can be ignored. Again from among the possible triplets, we have only (25, 205), (35, 305), and neither yields an integer k .

Finally, when a = 1, we have 10 > k > 5, giving 9, 7, and 6 as potential integer quotients. From the possible triplets, we have the following possible ( ab , a 0 b ) pairs: (15, 105), (14, 104), (16, 106), and (18, 108). From these, only ( 15 , 105) and ( 18 , 108) yield integer k s.

Summing: 450 + 45 + 18 + 15 = 528

q.e.f.

ab=10a+b a0b=100a+b the remainder of 100a+b divided by 10a+b is -9b. if b=0 of course the remainder is 0 now we cek that if b is not zero then ab is a faktor of -9b atau -9b=ab*k=10ka+bk or b(-9-k)=10ka for any k thus b multiple of 5 or -9-k multiple of 5 if b=5 then ab=15 or 45 since -9b=-45 if k=1 then -b=ka and if k=6 then -b=4a no solution satisfied it, if k>=11 impossible because a,b is a digit, if k=-4 then -5b=-40a thus 18 satisfy it, if k<=-9 impossible. so the sum of all ab is: 10+15+18+20+30+40+45+50+60+70+80+90=528

.100a+b=k(10a+b) 10(10-k)a=(k-1)b When b=0, then k=10, a=1,2,…,9 When b<>0, then consider the value of k: K=9, 10a=8b, so ab=45 K=8, 20a=7b K=7,30a=6b,so ab=15 K=6,40a=5b,so ab=18 K=5,50a=4b, no solution Sum=10+20+…+90+45+15+18=528

a0b=100a+b & ab=10a+b _ _ _ _ _ _ _ (1) As ab divides a0b, a0b=nab _ _ _ _ _ _ _ _ _ _ _ (2) (where n is positive integer)
From (1) & (2) ; 100a+b=n(10a+b) So, b/a=(100-10n)/(n-1) _ _ _ _ _ _ _ _ (3) As a & b are one digit numbers, 0<b/a<9 and from (3); 0<(100-10n)/(n-1)<9, By solving this, we'll find 5<n<10; by putting this values in equation (3) we will get b/a values and by using it properly we find all such possible values (ab).

Let's take a number 105. so, 105=100+15 ; 105= 90+15. Now factorize 90 and we get 10, 15 ,18 inside 10 - 19. Similarly 405= 360+45. Factorizing it we only get 45 as a divisor of 360 inside 40-49. so if we factorize 90(100-10), 180(200-20),.....810(900-90) and find their divisor between their region then we are done. So the answer is (10+20+30+..........+90) +15+18+45=450+15+18+45=528. And 528 is the answer.

All the pars that b = 0 satisfies the problem. So, the pars(a,b) = (1,0);(2,0)(3,0)(4,0)(5,0)(6,0)(7,0)(8,0)(9,0) are in the sum. And the pars (a,b) = (1,5);(1,8);(4,5) also are in the sum. So , 10+20+30+40+50+60+70+80+90 +15+18+45 = 528 is the sum of all 2 digit numbers N.

We split N to tow parts: Tens and Units. In Tens part for each i we have: i=10 till i<99 next i will be i=i+10 and to make j's as Unit we have j=0 till j<9 next j will be j = j+1, so we have N = i+j and X = i*10+j Then we apply the Divided Condition if remainder of X/N is Zero we are going to add the N in sum: sum = sum+N;

Obviously 10 | 100, 20 | 200, 30 | 300, ... 90 | 900

So 10, 20, 30, ... 90 are solutions to N.

Since ((ab)) divides ((a0b)), then (100a+b)/(10a+b) = c, for some integer c such that 0 < c < 9

By cross multiplying, and factoring, we get 10a(10 - c) = b(c - 1)

And we try to find different ways to factor 10, 20, 30, ... 90

10 = 2x5, 20 = 10x2 = 5x4, 30 = 5x6=10x3, 40 =5x8=10*4, 50=10x5

Comparing them to the equation above, we get (a,b,c) = (1,5,7), (1,8,6),(4,5,9)

So the answer is 10+20+30+40+50+60+70+80+90+15+18+45=528

Let $\overline{a0b} = k \overline{ab}$ for a positive integer $k$ . Clearly $k\leq 10$ as $10\overline{ab} = \overline{ab0} \geq \overline{a0b}$ . Then $100a + b = k(10a+b)$ $\Rightarrow 10(10-k)a=(k-1)b$ .

For $k=1, 2, 3, 4$ and $5$ we have $90a = 0, 80a = b, 35a = b, 20a = b$ and $25a = 2b$ , respectively. Clearly this has no solutions where $1\leq a \leq 9$ and $0 \leq b \leq 9$ .

For $k=6$ we have $8a = b$ . Since $b \le 9$ this implies $a=1$ is the only possibility. Thus $a = 1, b = 8 \Rightarrow N=18$ is the only solution for this case.

For $k=7$ we have $5a = b$ . Since $b \leq 9$ this implies that $a=1$ is the only possibility. Thus $a=1, b = 5 \Rightarrow N = 15$ is the only possible solution for this case.

For $k = 8$ we have $20a = 7b$ . Since $\gcd(7,20)=1$ this implies that $20$ must divide $b$ which gives $b=0$ . Thus $a=0$ , but $a \geq 1$ , hence there are no solutions for this case.

For $k=9$ we have $5a = 4b$ . Since $\gcd(4,5)=1$ this implies that $5$ divides $b$ which gives $b=5$ as the only possibility in the given range. Thus $a = 4$ . So $a=4, b=5 \Rightarrow N=45$ is the only possible solution for this case.

For $k=10$ we have $0a=9b \Rightarrow b = 0$ . Thus all $1\leq a\leq 9$ will satisfy the equation and we have $N=10,20,30,40,50,60,70,80,90$ as possible solutions for this case.

Hence the sum of all solutions is $18+15+45+10+20+30+40+50+60+70+80+90 = 528$ .

Here's how I did it: The question asks all numbers of the form $10a+b$ such that they divide the corresponding numbers of the form $100a+b$ .

Now, if that were to happen, $10*(10a+b)-(100a+b)$ should be divisible by $10a+b$ .

This means that $9b$ is divisible by $10a+b$ . For example, setting $b=0$ we get $9b=0$ which is divisible by all numbers of the form $10a$ where $a=1,2,3...9$ . Therefore $10,20,30,...90$ are some of the numbers.

Now setting $b=1$ we see that $9b=9$ has to be divisible by the numbers $10a+1$ . Clearly, this is impossible for any $a$ . We can similarly check the the result for other values of $b$ , and we find that the only other numbers that satisfy this are:

• $15$ and $45$ , for $b=5$ and
• $18$ , for $b=8$ . So the answer is $10+20+30+...90+15+18+45 = \boxed{528}$ .

If $(10a+b) | (100a+b)$ , then $(10a+b) | [(100a+b) - (10a+b)] \implies (10a+b) | (90a)$

Now there are two cases:

When $b = 0$ ,

$10a$ always divides $90a$ for any value of $a$ .

So, $\overline{ab} = 10, 20, 30, 40, 50, 60, 70, 80, 90$

When $b \neq 0$ ,

$10a+ b$ never divides $a$ but it can divide $90$ .

So, $\overline{ab} = 10, 15, 18, 30, 45, 90$

Therefore, our numbers are $10, 15, 18, 20, 30, 40, 45, 50, 60, 70, 80, 90$ which after adding gives $528$ .

if $ab\mid a0b$ and $b \ne 0$ then $\exists k \in \mathbb{Z} \because (10a+b)\times k = 100a+b$ \Rightarrow 10a(10-k) = b(k-1) \tag{1}

From which we can deduce that $2 \le k \le 9$ and $(k - 1, 10) \ne 1$
Moreover $\frac{10(10-k)}{k-1} < 10 \Rightarrow 110 < 20k \Rightarrow k > 5$ as $k \in \mathbb{Z}$

Thus $k \in {6, 7,9}$ , replacing k in (1) we get

For $k = 6, 10a(10-k) = b(k-1) \Rightarrow 10a\times 4 = b \times 5 \Rightarrow 8a = b$ so a = 1, b = 8

For $k = 7, 10a(10-k) = b(k-1) \Rightarrow 10a\times 3 = b \times 6 \Rightarrow 5a = b$ so a = 1, b = 5

For $k = 9, 10a(10-k) = b(k-1) \Rightarrow 10a\times 1 = b \times 8 \Rightarrow 5a = 4b$ so a = 4, b = 5 So the numbers are {15, 18, 45}\tag{2}

if b = 0, then $10a(10-k) = b(k-1) = 0$ as $a \ne 0$ we get $k = 10$ which would satisfy for all values of a i.e. $0 \le a \le 9$

So the numbers are {10, 20, 30, 40, 50, 60, 70, 80, 90}\tag{3}

Summing (2) and (3) we get 528

a0b¯¯¯¯¯.=(100 \times a)+(10 \times 0)+b=M(say) N=(10 \times a)+b Let, M/N=x(say) so, x=frac {100a+b}{10a+b}=an integer so, a \times (100-10 \times x)=b \times (x-1) so, \frac {10a}{b}=\frac {(x-1)}{(10-x)} Now if x=1, a=0 so it can't be possible.

For x=2, b=80 \cdot a.

For x=3, 3b=70 \cdot a.

For x=4, b=20 \cdot a.

For x=5, 2 \cdot b=25 \cdot a.

Now, a & b have the range of (0,9) So any of these is not possible.

Now for x=6 we get, b=8 \cdot a so we get a=1 & b=8 So, one of them is 18.

For x=7 we get, b=5 \cdot a So we get a=1 & b=5 So, another one is 15.

For x=8 we get, 7 \cdot b=20 \cdot a So,we can not get any exact value of a & b.

For x=9 we get, 4 \cdot b=5 \cdot a we get 1 pair of (a,b)=(4,5) So another one is 45.

Now if b=0 then we get, x=\frac {100a}{10a}=10 for this a can take the value from 1 to 9 i.e. the values are = 10,20,30,40,50,60,70,80 and 90.

So the summation of all the numbers is= 15+18+45+(10+20+30+40+50+60+70+80+90)=528

To make the units digit fit, the a0b must be divisible by 11,21,31,41,51,61,71,81,or 91. I found all numbers of the form a0b divisible by any of those numbers.