BdMO 2020 Secondary  Regional  P15

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 Joined:Sat Apr 03, 2021 7:41 pm
How many ordered pairs of integers $(a, b)$ are there such that $100 \leq a,b \leq 200$ and no carrying are required when calculating $a + b ?$
Last edited by tanmoy on Fri Apr 09, 2021 2:27 am, edited 1 time in total.
 gwimmy(abid)
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 Joined:Tue Apr 06, 2021 11:23 am
Re: BdMO 2020 Secondary  Regional  P15
Case 1: $a,b<200$
In ten's place, the order of the numbers matter, and we are limited to $9$ from $0$ as our sum. The ordered pairs for summation of $9$ is $(0,9), (1,8),...(4,5)$. In general, we can say that the number of ordered pairs for getting $a$ as sum is,
$$\begin{align*} n(a) =\begin{cases} \frac{a}{2}+1 & \text{when } a \text{ is even} \\ \frac{a+1}{2} & \text{when } a \text{ is odd} \end{cases}\end{align*}$$
The reason for it is, the ordered pairs start at $0$ and end at $\frac{a}{2}$ rounded down to nearest integer.
Counting $$\sum_{a=0}^9 n(a)$$ we get $30$
Ordering doesn't matter for the numbers in unit place. And there are $55$ total ways to form sums from $0$ to $9$.
Therefore, total possible pairs is $55\cdot 30$
However $a,b$ are distinct integers so we need to subtract $100$ from our result, since that is the total number of cases where $a=b$
Case 2: taking 200 into account
Any numbers can form a sum with $200$ and we are limited to 100 such numbers. thus we need to add $100$ to our result.
Thus our final number of ordered pairs is $55\cdot 30 = 1650$
In ten's place, the order of the numbers matter, and we are limited to $9$ from $0$ as our sum. The ordered pairs for summation of $9$ is $(0,9), (1,8),...(4,5)$. In general, we can say that the number of ordered pairs for getting $a$ as sum is,
$$\begin{align*} n(a) =\begin{cases} \frac{a}{2}+1 & \text{when } a \text{ is even} \\ \frac{a+1}{2} & \text{when } a \text{ is odd} \end{cases}\end{align*}$$
The reason for it is, the ordered pairs start at $0$ and end at $\frac{a}{2}$ rounded down to nearest integer.
Counting $$\sum_{a=0}^9 n(a)$$ we get $30$
Ordering doesn't matter for the numbers in unit place. And there are $55$ total ways to form sums from $0$ to $9$.
Therefore, total possible pairs is $55\cdot 30$
However $a,b$ are distinct integers so we need to subtract $100$ from our result, since that is the total number of cases where $a=b$
Case 2: taking 200 into account
Any numbers can form a sum with $200$ and we are limited to 100 such numbers. thus we need to add $100$ to our result.
Thus our final number of ordered pairs is $55\cdot 30 = 1650$
Umm....the healer needs healing...
Re: BdMO 2020 Secondary  Regional  P15
I also tried.
My total ans is = 4*55*55 = 12100
My total ans is = 4*55*55 = 12100