Michael's number

Let positive integers $a,b,c,d < 10$ and $a>0$

$M=\overline{abcd} = 1000a+100b+10c+d$

$T(M) =1000b+100a+10d+c$

$T(M)=4M$

$\Rightarrow 1000b+100a+10d+c = 4000a+400b+40c+4d$

$\Rightarrow 600b + 6d = 3900a+39c$

$\Rightarrow 200b + 2d = 1300a +13c$

$\Rightarrow 2(100b+d) = 13(100a+c)$

$\Rightarrow 100b+d = 13(50a+\frac{c}{2}$ )

Since $100b+d$ is a 3 digit number:

$50a+\frac{c}{2} <100$

$c<10$ hence $50 \leq 50a +\frac{c}{2} \leq 54$

$100b+d$ has middle digit $0$

Checking the options, we see that $50a+\frac{c}{2} =54 \Leftrightarrow 702=13 \times 54$

Giving $a=1 ,b=7 ,c=8, d=2$

$M=1782$ so the last three digits are $\fbox{782}$

$M$ is a four-digit integer so it can be written as: $M = \overline{abcd}^{(10)}$

And by definition of $T(M)$ : $T(M) = \overline{badc}^{(10)}$

Now, since $4M = T(M)$ , we have:

$4a\cdot10^3+4b\cdot10^2+4c\cdot10+4d = b\cdot10^3+a\cdot10^2+d\cdot10+c$

Wich gives after simplification: $1300a - 200b + 13c - 2d = 0$

Since $T(M)$ is at most a four-digit integer (can be a three-digit one if $b=0$ ), it has to be lower than $10,000$ . And since $4M = T(M)$ , we have:

$4M < 10,000 \Rightarrow M < 2,500 \Rightarrow M \leq 2,499$

This means that $a\in\{1,2\}$ .

However, if $a=2$ , then with the possibility of having a carry out when multiplying $M$ by $4$ , this gives:

$b\geq4a \Rightarrow b\geq8 \Rightarrow 4b\geq32$

And thus: $4M = 4\times(a\cdot10^3+b\cdot10^2+c\cdot10+d) \geq 4\times(a\cdot10^3+b\cdot10^2)$

$\Rightarrow 4M \geq 4a\cdot10^3+4b\cdot10^2 \geq 8\cdot10^3+32\cdot10^2$

$\Rightarrow 4M \geq 11.2\cdot10^3 > 10,000 > T(M)$

We arrive at a contradiction, so $a\neq 2$ . Hence $a=1$ .

Furthermore since there can be no carry out on the lower digit $d$ , we know that: $c = 4d$ .

By replacing these in the previous equation, we get:

$1300 - 200b + 50d = 0 \Rightarrow 130 - 20b + 5d = 0$

$\Rightarrow b = \dfrac{130 + 5d}{20}$

Moreover, since $c < 10$ and $c = 4d$ , we have $d\leq2$ and thus: $\left\{\begin{array}{ccc}c &\in& \{0,4,8\} \\ d &\in& \{0,1,2\}\end{array}\right.$

• if $d=0$ , then $b = \frac{130}{20} \notin \mathbb{N}$
• if $d=1$ , then $b = \frac{135}{20} \notin \mathbb{N}$
• if $d=2$ , then $b = \frac{140}{20} = 7$

Conclusion: $\left\{\begin{array}{cccc}a &=& 1 & \\ b &=& 7 & \\ c &=& 8 & (= 4\times2) \\ d &=& 2 &\end{array}\right.$

Hence: $M = 1782$ and $T(M) = 7128 = 4\times1782$

The answer is $782$

Let the number be $\overline{abcd}$ . So by condition we have, $4 * \overline{abcd}=\overline{badc}$

or, equivalently on expanding, $13(100a+c)=2(100b+d)$ .

So $13 | (100b+d)$ . Testing values of 3 digit multiples of 13 where ten's digit is 0, & testing if $\frac{2}{13} (100b+d)$ also has ten's digit 0, we get a unique favorable case for $a=1;b=7;c=8;d=2$ . Thus ans is 1782 ,easily verifiable.

In order to represent a 4 digit integer with variables, we use the following form:

1000a + 100b + 10c + d

Where a,b,c and d are the digits of the integer, from left to right respectively. Also, the following limitations are placed on these values:

0 <= a < 10|a is an integer 0 <= b < 10|b is an integer 0 <= c < 10|c is an integer 0 <= d < 10|d is an integer

Now, since T(M) = 1000b + 100a + 10d + c, the following is what we wish to find:

1000b + 100a + 10d + c = 4(1000a + 100b + 10c + 1d) 1000b + 100a + 10d + c = 4000a + 400b + 40c + 4d -3900a + 600b - 39c + 6d = 0 -1300a + 200b - 13c + 2d = 0 200b - 13c + 2d = 1300a

200b - 13c + 2d = 1300a

We know that a has to equal at least 1 since the number is at least 4 digits long. The maximum possible value of 200b - 13c + 2d, following the previous restrictions, is 1818. Therefore a must be 1, since any value larger than 1 would results in the right side equalling 2600, which makes the equality impossible.

If a is one, we can substitute it back into the equation as follows:

200b - 13c + 2d = 1300

Since 200b + 2d - 13c has to equal 1300, then b must equal at least 7. It it were smaller than 7, for example 6, then the equation would simplify to 2d = 13c + 100, and since the maximum limit on 2d is 18, this equation would be impossible to simplify. Any smaller inputs of b would simply make the right side bigger, making it even more impossible! Similarly, if b was larger than 7, for example 8, then the equation would be 2d + 300 = 13c. Since the maximum limit of 13c is 117, then this equation is impossible to satisfy, and any values of b higher than 7 would simply increase the left side further! Therefore, b = 7.

Substituting it back into the equation, we get the following:

2d + 100 = 13c

Since 2d + 100 will always be an even number larger than 100, c must be an even number which, when multiplied with 13, is larger than 100, but still belongs to the realm of 0 to 1, inclusive. The only value that is possible is 8, which when substituted into the equation gives us:

2d + 100 = 104 d = 2

Therefore the number M is 1783.

the units digit of M can be 1 or 2 => the tens digit of M can be 4 or 8. Let the thousands and hundreds digits of M be a and b respectively. We have 4x(1000a + 100b + 40 +1) = 1000b + 100a + 10 + 4, or, 4x(1000a + 100b + 80 + 2) = 1000b + 100a + 20 + 8. The first equation doesn't give us any viable solutions. The second one gives us a=1 and b=7, thus the number is 1782.

Let the number be $\overline{abcd}$ . According to the question:

$4\overline{abcd} = \overline{badc}$

$\implies4(1000a + 100b + 10c + d) = 1000b + 100a + 10d + c$

$\implies 39(100a + c) = 6(100b + d)$

$\implies 39\overline{a0c} = 6\overline{b0d} \implies 39\overline{a0c}/6 = \overline{b0d} \space\space \space (1)$

Clearly, $\overline{b0d} \leq 909 \implies \overline{a0c} \leq 909\times6/39 = 139.84$

Also, $6|\overline{a0c}$ (from $(1)$ . The only two numbers that satisfy the criteria are $104, 108$ . Out of these $39\times 104/6 = 676$ , which clearly is not of the required form. $108\times 39/6 = 702$ .

Therefore, $a = 1, b = 7, c = 8, d = 2$ and the number equals $\boxed{1782}$ .

We are told that if $M=1000a+100b+10c+d$ then $4000a+400b+40c+4d=1000b+100a+10d+c$ .

Therefore $1300a-200b+13c-2d=0$ . $*$

Considering this equation $mod 100$ , $13c-2d \equiv 0$ .

By inspection of even values of $c, c=8$ and $d=2$ are the only possibilities.

Substituting back into $*$ , $1300a-200b+300=0$ , giving $13a-2b+3=0$ .

By inspection of odd values of $a, a=1$ and $b=7$ are the only feasible solutions.

Therefore $M=1782$ .

Let M=1000a+100b+10c+d where a,b,c,d are whole nos. from 0 to 9

here a,b is not = 0 (as this would lead to 3 digit no.)

T(M)=1000b+100a+10d+c

now T(M)=4M

therefore 1000b+100a+10d+c=4000a+400b+40c+4d

solving this we get

1300a+13c=200b+2d

now here a can't be 2 or above (as this would lead to b>9 which is not possible)

therefore a=1

now by hit and trial in the equation 1300+13c=200b+2d

b=7, c=8, d=2 (remember b,c,d are integers from 0-9)

therefore M=1782

last three digits of M are 782

Suppose $M = 1000a + 100b + 10c + d$ , where $a,b,c,d$ are digits, then we are given $4000a + 400b + 40c + 4d = 1000b + 100a + 10d + c$ This can be written as $3900 a-600 b+39 c-6 d = 0$ from this $c$ is even, hence $c = 2x$ , substituting this and dividing by $6$ we get $d = 650a - 100b + 13x$ Remember that $0 \leq a,b,d \leq 9$ and $0 \leq x \leq 4$ . So try $x = 4$ (so that $d$ is still a digit). From this it follows that $(a,b,c,d) = (1,7,8,2)$ , checking we get that it is indeed a solution, we are given that the solution is unique, hence we are done, and the last three digits are $\boxed{782}$