What's My Number?

Consider the string 9876543210. We essentially have to take out two digits so that the resulting number is divisible by 180 = 2X5X2X3X3. Now the number is divisible by 10, so clearly the last digit stays 0. The number is also divisible by four; a number ending in 10 is not divisible by 4, but a number ending in 20 is; therefore we remove the 1. The desired number is also divisible by 9, so the sum of its digits is divisible by nine. Having subtracted 1, we also must remove 8 to satisfy the condition. So Calvin's number is 97654320, and the first three digits are 976.

Yeah , this question is a nice one :) .

Though you'll find find my solution pretty huge , but then too . It is extremely explanatory and includes whatever went through my mind and involves all the mental logic that I applied . So here it is :

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*The Beginning: *

It is given that $\mathbf{N}$ is divisible by $\mathbf{180}$ .

Thus, divisors of $\mathbf{180}$ and $\mathbf{N}$ will be same .

This follows that $\mathbf{N}$ is divisible by $2,4,5,9,10 \dots$ etc.

This point follows that [From the properties of divisibility] :

1. Since $10|N$ , The last digit will be $\mathbf{0}$ (zero) .

2. Since $4|N$ , The number formed by the last two digit will be divisible by 4 .

3. Since $$ $9|N$ , The sum of the digits of $\mathbf{N}$ is divisible by $9$ .

Okay . We are all set to find Calvin's favorite number $\mathbf{N}$ .

$$

*The Number : *

So we fill the values we have found .

Suppose the number is $\overline{ABCDEFGH}$

The number we get from (1) is :

$\boxed{A} \boxed{B} \boxed{C} \boxed{D} \boxed{E} \boxed{F} \boxed{G} \boxed{0}$

$$

*Let the Arguments Begin : *

Now since the number is strictly decreasing ,

Thus , maximum value of $G$ will be $3$ as $A$ cannot be greater than $9$ .

[ Think over it. If we take $G=4$ we'll require at least $A = 10$ which is simply not possible ]

The above argument and (2) implies that we'll have to find a number of the form $\overline{G0}$ which is a multiple of $4$ and $G \leq 3$ .

Which is simply : $20$ . Hence , $G=2$

Thus , the number now becomes :

$\boxed{A} \boxed{B} \boxed{C} \boxed{D} \boxed{E} \boxed{F} \boxed{2} \boxed{0}$

As the number is strictly decreasing , there won't be any digit equal to $1$ . [ Think :) ]

$$

Towards the final number:

Finally , according to the third point ,

I paired up numbers that add up to $9$ i.e. $(0,9) , (1,8) , (2,7) , (3,6) , (4,5)$ .

[As $A,BC,D,E,F,G,H$ all are single digit numbers they'll add up to $9$ only]

Here, we exclude $(1,8)$ , as it won't help us in finding $N$ $$ [ Think over it. If we include $8$ , neither of the digits of $N$ when added to $8$ will give us $9$ as we have already confirmed that $1$ will not exist in $N$ ]

Our argument above satisfies that including $8$ is not an option .

$$

*The Final Number : *

So we'll have $6$ spaces i.e $ABCDEF$ and $6$ numbers i.e $9,7,6,5,4,3$ .

Now since - order is decreasing , we arrange these numbers to get :

$\boxed{9} \boxed{7} \boxed{6} \boxed{5} \boxed{4} \boxed{3} \boxed{2} \boxed{0}$

Which is the only possible combination .

Thus the answer is $\overline{ABC}$ i.e $\boxed{\mathbf{976}}$

Cheers !

Whoa! what a big number I wonder if it is his password ;)

We know that the last digit has to be 0 or else it wound not be divisible by 180 (because of 10 in its factor)

Now if the last digit is 0, then we cancel 0 of 180 and the 8 digit number

Now we have a 7 digit number which should be divisible by 18

The last digit must be even i.e. 2 (it cannot be 1)

Now we have 6 digits which should be made by remaining 7 digits (removing 0,1and 2)

That is made from 9,8,7,6,5,4,3,

The number will be divisible by 9 when the sum of all digits is divisible by 9

From this we get 9+7+6+5+4+3+2=36

The number is 97654320

Let the number $N$ be $\overline{abcdefgh}$ . From the first and second conditions given in the question, we have, $a>b>c>d>e>f>g>h$ . Now to satisfy the third condition, we need a little bit elaboration. So lets get started:

1. We know that $180=2^2*3^2*5=9*20=10*18=4*45$
2. Now, as N is divisible by $180$ , it must also be divisible by $9$ , $10$ , and $4$ . We're gonna use the divisibility tests for these numbers to solve the problem.
3. For a number to be divisible by $9$ the sum of its digits must also be divisible by $9$ . Note that the maximum digit sum of $N$ occurs when $a,b............,g,h$ are $9,8.....3,2$ respectively. Hence $N's$ maximum digit sum is $44$ . $N's$ minimum digit sum, which is equal to $28$ ,occurs when $a,b......g,h$ are $7,6.............1,0$ respectively. There is only one multiple of $9$ between these limits. Thus, it follows that $a+b+c+d+e+f+g+h=36$
4. Because $N$ is multiple of $10$ , hence it unit's digit, that is $h$ must be zero. Hence $a+b+c+d+e+f+g=36$
5. Lastly, because $N$ is a multiple of $4$ the last two digits of $N$ (which are $\overline{gh}$ ) must be divisible by four. As $h=0$ , the digit $g$ could be $2,4,6$ or $8$ . With $g=2$ as the only possibility (because if $g$ is equal to $4,6$ or $8$ , $a$ would have to be greater than $9$ , which is not possible), we have $a+b+c+d+e+f=34$
6. Now we have minimum value of $f$ as $3$ . Note that if $f>3$ , then the minimum (and the only possible) value of the sum of these digits must be $39$ , which exceeds $34$ . Hence, the only possible value of $f$ is $3$ . Hence $a+b+c+d+e=31$ .
7. Similarly, we can prove that $e=4,d=5$ and $c=6$ . So we are left with the equation $a+b=16$ . With $a=9$ and $b=7$ as the only possibility of distinct digits $a$ and $b$ such that $a>b$ , we have $a=9$ , $b=7$ and $c=6$ . Hence $N=97654320$ our answer is $\overline{abc}$ = $\boxed{976}$

180=2x2x3x3x5... It is obvious the last digit is zero as it is divisible by 180....... the final 0 gets rid of a 2 and a 5 from the prime factors...... now 18=9x2 and so the remaining digits must have a sum divisible by 9 and the last digit before 0 must be 2...as there are 7 ascending digits.... so the options are from 2 to 9 with one digit excluded... 2+3+4......+9=44 therefore the nearest sum divisible by 9 is 36... hence 8 must be cut therefore the first 3 nos are 976

180=2 2 3 3 5 so required number should be divisible by 2 3 and 5 so number must be even and last digit 0 for divisibility by thhree sum of digits divisible by 3 so number is 97654320

First of all, since N is divisible by 180 (2 2 3 3 5), it is divisible by 3 3 = 9, and 2 2*5 = 20. Therefore the ones digit must be zero and the tens digit must be even. Since the digits are increasing from the ones digit, the tens digit must be 2 or else we will run out of numbers. Also, the rest of the 6 digits must add to a multiple of 9.

So far N looks like:ABCDEF20.

If the digits increases by 1 from 0, then N would be 87654320. However the sum of the digits of N are 2+3+4+5+6+7+8+0 = 35 which is not divisible by a multiple of 9. Because the sum cannot decrease from 27 since 87654320 is the lowest possible value that is divisible by 20, the sum of the digits must equal 36 (the next multiple of 9).

Therefore, once the sum of the digits of 87654320 is increased by 1, the value of N is 97654320 which is divisible by 180. Therefore the first 3 digits of N are 976.

Here we are given an 8 digit number with distinct digits, divisible by 180, So the last digit is 0 and the penultimate digit is even, and hence is true. Hence we have numbers from 9-3 to be filled in 6 places. As the number is divisible by 180, the sum of its digits should be a multiple of 9. And thus we remove 8 from the remaining numbers and get N = \boxed{97654320} and hence the first three digits are \boxed{976}

Since N has exactly 8 distinct digits, it is composed of the digits 0-9 with 2 missing. Thus, N can be thought of as 9876543210 with 2 digits taken out (and divisible by 180).

Since N is divisible by 180, it must be divisible by both 20 and 9. Since N is divisible by 20, the last two digits must be 20, 40, etc. For the last two digits to be 20, you take out the single 1, but for the last two digits to be 40 or greater, you must take out at least 3 digits (if it ends in 40, it cannot have the digits 3, 2, or 1, and similarly for 60 and 80). Thus, the last two digits are 20.

Updated, N must be 987654320 and divisible by 9.

A trick for divisibility-by-9 is that the sum of the digits must be divisible by 9. Note that 9876543210 is divisible by 9 because the sum of each "shell" (9+0, 8+1, 7+2, 6+3, etc.) is divisible by 9. We have already taken out the digit 1, and if we take out the digit 8 for our second and last digit, then then the entire shell would be taken out (not changing the divisibility of the sum of the digits, as 9k-9 is still divisible by 9).*

This gives N = 97654320 and the answer $\boxed{976}$ .

*Inversely, if a number other than the 8 is taken out along with the 1 (for example the 5), then two incomplete half-shells would remain (in this example, 8 and 4). The other 3 shells (9+0, 7+2, 6+3) would still have sums of 9, and so the divisibility of the eight-digit number must be the same as the divisibility of the two-digit number formed by the two halves of the incomplete shells (8+4). However, because the only digit you can add to 8 to make a multiple of 9 is 1 (and the numbers 10, 19, 28, etc. but those aren't digits), the other half-shell (4) cannot be added to 8 to make a multiple of 9 (8+4=12 is not a multiple of 9).

as we may know, 180's greatest multiple that is less than 1000000000 is 99999900, that is equal to :

180x555555

so to find a strictly decreasing digit, we'll have to subtract at least 1800000 (equals with 180x100000) with it so the digit would not be the same

thus we obtained

99999900-1800000 = 9819990

because we are not allowed to place 1 before 9 , we need to subtract at least 540000 (equals with 180x3000) so we will obtain

9819990-540000=97659900

the properties are almost fulfilled , but we still need to subtract 5400 (equals 180x3) to change the "9900" at the last four digits thus we obtain again

97659900-5400= 97654500

to change the "500" at the last three digits , we will need to subtract 180 (equals to 180x1) thus we obtain

97654500-180=97654320

so we obtained the first three digits are 976