Mystery Four-Digit Number

The first digit is one-third the second last is three times the second indicates that the second term is divisible by 3 but < 10. so, the number can be 3, 6, or 9. But, last is three times the second also means that the last digit can only be 3 3=9, 6 3=18 or 9*3=27. But the last digit must be < 10 Hence, last digit = 9, second digit = 9/3 = 3 first digit = 3/3 = 1 third digit = 1 + 3 = 4 Putting that together you get 1349

a b c d // 1a 3a c d // 1a 3a 4a d // 1a 3a 4a 9a

Let t number be A B C D

Step 1. A=B/3, => B=3 A
Step 2. C=A+B, => C=A+3 A, => C=4 A
Step 3. D=3 B, => D=3(3 A), => D=9 A

Therefore A B C D => 1A 3A 4A 9A

The first condition implies that the second term is a multiple of 3. The last condition, i.e., "the last is three times the second" tells us that the second digit MUST be 3 because 3*3=9, which occupies the unit place. So the first digit is 1/3(3) = 1, second digit is 3, third digit is 1+3=4 and the fourth digit is 9. Hence the number is 1349.

The 1st digit is a third of the 2nd, creating an automatic ratio in your head: 1 3 The 3rd is the SUM of the 1st and 2nd: 1+3=4 So far the number is: 1 3 4

Finally, the 4th digit is three times the 2nd: 3x3=9 So the end number is: 1 3 4 9

Let the first no. Be 'x'. So the second would be 3x. Similarly the third, x+3x, would be 4x. Last one, obviously, is 9x. The no. is 'x 3x 4x 9x'. To get the exact no., substitute x as 1. =>1349

Let the second digit be x. First digit- x/3 Second digit- x Third digit- x+(x/3) Fourth digit- 3x So the number has to be divisible by 3, 6 and 9 as it has to be less than 10. Also its just a 4 digit no. So the multiple 3x should not be more than 10. Hence the x=3. So the no. Is 1349.

here mentioned digit....possible second digit can only 3, 6 or 9....so first two digit of no can be 13, 26, 39....now third digit is sum of first and second digit so 134, 268 only possible value....39 is not possible because (3+9=12) which is not single digit.....now last term can be only be 9 because (6*3=18) not possible....so ans 1349

Let the second digit be n. The first digit is 1/3 of the second digit so the first digit is 1/3 n. The third digit is the first added to the second so it is n+1/3n = 4/3 n. The fourth digit is three times the second so it is equal to 3n. Therefore the digits are: 1/3 n, n, 4/3 n, 3n. 1/3n has to be a whole number so n is a multiple of 3. 3n has to be <10 because it is a single digit. Therefore n<10/3. As n is a positive multiple of 3 less than 10/3, n must equal 3 or 0. In this case n can't equal 0 because that would make the number 0000 which is not a four digit number, so n has to be 3. This means that the digits are: (1/3 x 3), 3, (4/3 x 3), (3x3) = 1349

let 2nd digit be x
1st digit=1/3 of x
=x/3
3rd digit=x+x/3
=4x/3
4th digit=3x
So the no. is (3x 1000)+(4x/3 *100)+(x 10)+(x/3*1)
Since all the digits are non fraction integers , the only value of x is 3

I used code to solve this: don't mind the first three lines of output; its the fourth one

solve it by hit n trial method

it can be written as 1000 x +100(3 x)+10(4 x)+9x where 3x is the second digit

It is given in the question that

$a = \frac {b}{3}$

$c = a + b = \frac {4b}{3}$

$d = 3b$

For the number denoted as $\overline {abcd}$

The highest value (in terms of $b$ ) is $d$ . None of the numbers can exceed ten so

$d < 10 \therefore 3b < 10$

So the maximum number $b$ can be (while still being an integer) is $3$ so let's make $b$ equal to $3$

$a = \frac {(3)}{3} = 1$

$b = 3$

$c = \frac {4(3)}{3} = 4$

$d = 3(3) = 9$

We've been told there is only one four-digit number satisfying the conditions so the answer is

$\overline {abcd} = \boxed {1349}$

First digit=(x/3) second digit=x third digit=x+(x/3) fourth digit=3x put x=3 then answer is 1349

start 1st no. from 1 and get answer

let x/3, x, x+x/3, 3x be the 4 digits, now substitute the least possible value of x so that u end up with an integer at x/3

Let number be $1000a+100b+10c+d$ It is given that: $a=\frac{b}{3}\rightarrow\;3a=b$ $c=a+b$ $d=3b$ Putting $3a=b$ in the equation for $c$ gives $c=a+3a=4a$ .If $3a=b$ then $d=3b=3(3a)=9a$ .So $1000a+100b+10c+d=1000a+100(3a)+10(4a)+(9a)=1000a+300a+40a+9a=1349a$ So the number is $\boxed{1349}$

let the digits be a,b,c,d. By given data b=3a, c=4a, d=9a. put a=1 (why - coz if 2 is substituted d's value will be greater than 10. but "a single digit" cannot be 10)... thus we get 1349

let the no. be abcd a=b/3 c=a+b=4b/3 d=3b b must be a multiple of 3 b= 3,6,9 but d must be less than 10 3b<10 b<3.33 b=3 a=1 c=4 d=9 required no. is 1349

Let the second digit be x. First digit- x/3 Second digit- x Third digit- x+(x/3) Fourth digit- 3x So the number has to be divisible by 3, 6 and 9 as it has to be less than 10. Also its just a 4 digit no. So the multiple 3x should not be more than 10. Hence the x=3. So the no. Is 1349.

The last digit has to be 3 times the 2nd number, so the 2nd number need to be > or = 3. 1st number is 1/3 is 2nd, so 2nd has to be divisible by 3, so it is 3. 1/3 of 3=1. 1+3=4. 3x3=9. 1349

The first digit must be a mutliple of 3, and smaller than 10. That leaves us with the possible results of 3, 6 and 9. The last digit is 3 times the second, so it must be 3 3, 3 6 or 3 9(a.k.a. 9, 18, or 27) and be smaller than 10. This leaves us with 3 3, or 9. The second digit will be 3, and the first is 1, 1+3=4, so the entire number is 1349

"The first digit is one-third the second[...]" => this means our number looks like this: a 3a 4a 9a => the only value possible for a is 1, as any other value will end up in a result with two digits for the last position of the number.

4th digit is 3 times the 2nd, and 2nd is 3 times the 1st digit. Therefore, the 4th is 9 times the first digit. As any digit can not be more than 9, therefore, the fourth digit being a multiple of 9, can only be 9. This means the 1st digit is 1 and the 2nd is 3. The 3rd digit being a sum of 1st and 2nd digits will be 1+3=4. Therefore, the answer is 1349.

the 4th number equals second number multiplied by 3 and the second number is three times the first which leads to the 4th number equals the first one multiplied by 9 and for sure both numbers are one digit less than 10 that means first one is 1 and the 4th is 9 second equals 1/3 the 4th = 3 and the third equals first +second 1+3 =4 then number is 1349

the first digit is one third of second one then consider second digit 3 & fist digit is 1.then third digit is sum of first and second digit then third digit is 4. and last digit is three times the second digit then fourth digit is 9 . Answer is 1349

Form the question, we get that that the 2nd digit is a multiple of 3<10 (so that you get a whole number in the thousands place). The only multiple of 3 which will satisfy the given rules is no. 3 itself. And if you compute according to the rules given, you will get the required number, 1349.

Call the number $ABCD$ .

From the first statement:

$A=\frac{1}{3}B$

The digits must be integers, therefore:

$A<B$

and

$B\mod3\equiv 0\mod3$

Therefore, the only choices for $B$ are: $3,6$ and $9$ .

Let $B=9$ . Then $A=3$ .

The second statement claims that:

$C=A+B\\C=9+3=12$

This is not possible in decimal, so $B\neq9$ .

Let $B=6$ . Then, $A=2$ .

The third statement claims that:

$D=3\cdot B \\D=3\cdot6\\D=18$

This is not possible in decimal, so $B\neq6$ .

The only remaning option is $B=3$ . Then, $A=1$ .

Verify that this works with the second and third statements, and if it works, then the seconds and third statements will give the value of $C$ and of $D$ :

$C=1+3=4$

$D=3\cdot 3=9$

Therefore, the number is $1349$ .

second no is 3. the first digit is one-third the second is 1,the third is the sum of the first and second is 4,and the last is three times the second is 9.ans is 1349. assume second no is 3. 1/3=3/3 =1 3+1=4,3*3=9 1349

first wiill be 1 or2 or 3 second will be 3 or 6 or 9 third will be 4 or 5 or 8 or 6 or 7 or 8 or 9 last is 3 times of the first i.e it will be 3 or 6 or 9 hence the number is 1349

I literally just guessed the first number was 1, which would make the second 3, and the third 4. Since 3 times 3 is 9, there's obviously not another number that could satisfy all of the other conditions

The first digit is one -third the second 3/3=1 and the third is the sum of first & second 1+3=4and the last is three times the second if according to it last digit is three times multiple of second number 3*3=9 so thats why the answer is 1349

the second digit is three times the first and again the last is three times the second, its only possible with three numbers 1,3 and 9 and the second digit is 4. simply

1= 1/3 3 = 1st number.. hence 3 = 2nd number third number = 1st number + 2nd Number = 1 + 3 = 4 4th number = 3 * 2nd number = 3 3 = 9

(a/3)(a)(a+1/a)(3a) a is divisible by 3 , and 3a<10 then a=3 ,answer is1349.

We see that the first digit cannot be 2; the number will not end up to be 4 digits. If the first digit is 1, then we can from there, finding the solution of 1349

Let's say:

w = 4th digit,

z = 3rd digit,

y = 2nd digit

x = 1st digit...

the # will be: wzyx (I just reverse it.)

then, by the given data,

x = (1/3)y ---> 3x = y

z = x+y --> z = 3x + x = 4x

w = 3y ---> w = 3(3x) = 9x

then:

the number will be:

(9x)-(4x)-(3x)-(x)

the max. value of x is 1 because we are just forming 1 digit.

then,

(9x)-(4x)-(3x)-(x) ---> 9431

then,

the # is just reverse of 9431 which is 1349..

Final Answer: 1349.