Big Middle-Digit Numbers

Consider values of $b$ for $0 \le b \le 9$ . $b$ cannot be equal to $0$ because $a$ would have to be $0$ as well, which would not constitute a three-digit number. For $b=1$ , the values of $a$ are $1$ and $c$ are $0,1$ , which is $1\left(2\right)=2$ permutations in total. For $b=2$ , the values of $a$ and $c$ are $1,2$ and $0,1,2$ , which is $2\left(3\right)=6$ permutations in total. The number of permutations of $a$ and $c$ for each value of $b$ is $b\left(b+1\right)$ . To find the total number of permutations, just evaluate $\sum_{i=1}^9 i\left(i+1\right)$ .

First of all we established that $b = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ .

When $b = k$ , since $a \leq b$ then, $a = \{k, k-1, \ldots, 2, 1\}$ . On the other hand, since $c \leq b$ then, $c = \{k, k-1, \ldots, 2, 1, 0\}.$ Thus there are, $k \cdot 1 \cdot (k+1)$ , three digit numbers $N$ .

This implies that the number of three digit numbers is the sum of such $k \cdot 1 \cdot (k+1)$ , where $k=\{0, 1, 2, \ldots, 8, 9\}$ .

$\sum_{k=0}^9k \cdot 1 \cdot (k+1)=\sum_{k=0}^9(k^2+k)$ $=\sum_{k=0}^9k^2 + \sum_{k=0}^9k$ $=\{\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\}_{n=9}$ $=\frac{9\cdot10\cdot19}{6}+\frac{9\cdot10}{2}$ $=3\cdot5\cdot19+9\cdot5$ $=285+45$ $=330$

Hence, there are $330$ such $N$ satisfying the given condition.

We can apply casework on b. When b = 1, there is one possibility for a (1) and two for c (0,1). When b = 2, there are two possibilities for a (1, 2) and three for c (0, 1, 2). Similarly, when b = n with n \leq 9, there are n possibilities for a (1, 2, ..., n) and n+1 possibilities for c (0, 1, ..., n), for a total of n(n+1) ways. Summing up n(n+1) for n = 1 to n = 9, we get 330.

And then I guessed that 330 would be the right answer based on my thought process.

Let's start with b: b cannot be 0 (otherwise a would be 0 and it can't); if b=1 $\Rightarrow$ a=1 and c=0 or 1 (1 2 possibilities); if b=2 $\Rightarrow$ a=1 or 2 and c=0 or 1 or 2 (2 3 possibilities) and so on to the case in which b=9 and there are 9*10 possibilities.

Now we have to sum these products: we can sum them product by product, there is no problem, but we can use a formula.

These products are the doubles of the triangular numbers ( $t_n$ = $\frac {n(n+1)}{2}$ ) The sum is $\sum_{i=1}^9$ $n^2 + n$ = $\sum_{i=1}^9$ $n^2$ + $\sum_{i=1}^9$ $n$ = $\frac {n(n+1)(2n+1)}{6})$ + $\frac {n(n+1)}{2})$ = 285 + 45 = 330

This problem is equivalent to finding all three-digit numbers such no digit is larger than the middle one. To do so, the first few cases can be investigated in search of a pattern.

Suppose b = 1. Then a can only be 1, while c may be 0 or 1. Thus, there is 1 choice for a and 2 for c, resulting in a total of 1*2 = 2 possibilities.

Suppose b = 2. Then a can be 1 or 2, and c can be 0, 1, or 2, resulting in a total of 2*3 = 6 possibilities.

The pattern can be seen here. Whatever value b has, a can take on any value from 1 to b inclusive, so if b = n, a has n choices. Similarly, c can take on any of those choices, plus 0, so c has n+1 choices. Thus, the total number of possibilities is:

\sum_{i=1}^{9}n(n+1) = 330.

a>0, else N is two digits, as opposed to three. Therefore, b>0, in order to satisfy a \leq b .

Let b=1. There is 1 possible value of a that satisfies the inequalities: 1. There are 2 possible values of c that satisfy the inequalities: 0 and 1.

In general, for of b=n, there are n values for a and n+1 values for c. Therefore, there are n(n+1) ways of rearranging a and c for each value of b. This gives rise to a sequence of terms, from b=1 to b=9, of 2, 6, 12, 20, 30, 42, 56, 72, 90. The sum of this finite sequence is 330. Therefore, the answer is 330.

we look at the possible cases for b, and determine how many a's and c's satisfy the condition.

if b=0, no a's or c's will do, since the conditions imply a = c = 0, which does not give a three digit number

if b =1, we have 1 choice for a (which is 1), and 2 for c (which are 0,1). so that's 1x2. if b =2, we have 2 for a, and 3 for c. 2x3. if we continue until b=9, we get the total:

1x2+2x3+...+8x9 = 330

this sum can easily be computed if we see that the sum above is the sum of all k(k+1) from k =1 to k=9 and use the summation formulas for sums of natural numbers and squares of natural numbers

given condition- "a" less than or equal to'b', 'c' less than or equal to 'b' so, the numbers will form a series- between 100 and 200-{110,111,120,121,122,130,131,132,133.......191,192,193,194,195,196,197,199} so,there are 54 numbers between 100 and 200 which satisfy the condition. then between 200 and 300 there will be 54 -2=52(since 210,211 are not possible as in case of 110,111)then between 300 and 400 there will be 54 -5 =49 numbers(same reason) like this goes on upto between 900 and 1000,there will be 54-44=10 numbers on counting these we will get 54+52+49+45+40+34+27+19+10 = 330

for any value of b = n a can have values [ 1,n ] i.e. n values and c can have values [ 0,n ] i.e. {n+1} values and as a\geq b so b can have a minimum value of 1 and a maximum value of 9 so ans = \displaystyle \sum_{i = 1}^9 *{i (i+1)} == 330

IF WE DO THIS BY CHECKING WHETHER ALL VALUES ARE OBLIGING TO THE GIVEN DATA WE GET 54 VALUES IN 100-200 RANGE. ALSO WE GET 52 VALUES IN 200-300 RANGE.

300-400 RANGE 49 VALUES 400-500 = 45 500-600 = 40 600-700 = 34 700-800 =27 800-900 = 19 900-999 =10

IF WE ADD ALL THESE VALUES WE WILL GET 330 NUMBERS

answer is 1(2)+2(3)+3(4)+4(5)+5(6)+6(7)+7(8)+8(9)+9(10)=330

Solution 1: Since $N = \overline{abc}$ , we must have $1 \leq a \leq 9$ , $0 \leq b \leq 9$ and $0 \leq c \leq 9$ . For $b = 0$ there are no such numbers that satisfy the given conditions in the question. Thus $b \geq 1$ .

For each $1 \leq b \leq 9$ , there are $b$ possibilities for $a$ $(1,2,\ldots,b)$ and $b + 1$ possibilities for $c$ $(0,1,2\ldots , b)$ . Thus by the rule of product there are $b(b+1)$ such numbers for each $b$ .

Therefore, we have $0 + 1(2) + 2(3) + 3(4) + 4(5) + 5(6) + 6(7) + 7(8) + 8(9) + 9(10) = 330$ such numbers.

Note: We can show that $\sum_{i=1}^n i(i+1) = \sum_{i=1}^n 2 \cdot {i+1 \choose 2} = 2 \cdot {n+2 \choose 3} = \frac {n(n+1)(n+2)}{3}$ , which shows that the sum is equal to $\frac {9 \times 10 \times 11 }{3} = 330$ .

Solution 2: First, let's ignore the restriction that $a\neq 0$ . By considering the value of $b$ , when $b=n$ , there are $(n+1) ^2$ possibilities for the pair $(a,c)$ . Hence, there are $1^2 + 2^2 + 3^2 + \ldots + 10^2 = 385$ possibilities in total.

We now count the number of times when $a = 0$ . By considering the value of $b$ , when $b=n$ , there are $(n+1)$ possibilities for $c$ . Hence, there are $1+2+3 + \ldots + 10 = 55$ possibilities in all.

Thus, there are $385 - 55 = 330$ total cases.

Between 100 and 200 the possible nos. are 100,110,111,120,121,122,130,131,132,133,140,141,142,143,144,150,151,152,153,154,155............ Take 100 - a=2 is not possible. So only 1 possibility Take 110 - In this also only 1 possibility Take 120- In this 220, 221 and 222 is possible. Like this it goes on..... So we can write it as 1+2+3x2+4x3+5x4+....10x9 which gives 330