Technothlon 2014 Question 9

See that both the numbers 1041 and 2221 have $16$ choices each to be a number in decimal representation, as each of its digits is corresponding to 2 digits in decimal.

But, if the first number starts with $6$ , then it's square becomes 8-digit, so that is not valid (we want square to be 7 digit number), so it starts with $1$ .

Similarly for the second number, it can't start with $7$ , so it starts with $2$ .

So we got $8$ numbers to check for 1st number and $8$ for the second.

The maximum number such that second digit of first number is 0 is $1096$ , and it's square is $1201216$ which is not having a $2$ at start in technometric representation, so the ones smaller than it will also not have.

Thus we have to check only 4 cases for the first number, namely starting with $15$

$\begin{array}{|c|c||c|} \textbf{number} & \textbf{square} & \textbf{technometry. .represent} \\ \hline 1541&2374681&2324131 (true)\\ 1546&2390116&2340111\\ 1591&2531281&2031231\\ 1596&2547216&2042211\\ \end{array}$

For the second number, there are $8$ choices similar way, namely

$\begin{array}{|c|c||c|} \textbf{number} & \textbf{square} & \textbf{technometry. .represent} \\ \hline 2221&4932841&(not- start-with -2)\\ 2226&4955076&(not- start-with -2)\\ 2271&5157441&(not- start-with -2)\\ 2276&5180176&(not- start-with -2)\\ 2721&7403841&24*****(no-use)\\ 2726&7431076&24*****(no-use)\\ 2771&7678441&21*****(no-use)\\ 2776&must-be-the-needed&2201121\\ \end{array}$

Thus $A=1541$ and $B=2776$ in decimal, giving $B-A = 1235$

And in Technometry, it will be $\boxed{1230}$

Technometics number is always less than or equal to 10 base normal number. $Also~ note~ that~ a~( X_{Digit}~ number )^2 = 2X_{Digit}~ or ~(2x-1)_{Digits}~number.\\ It~ will~ have (2x-1)_{Digits}~ if~ the~ highest~ digit~ is~ less~ than~~ 3.2. \\~\\$ $\displaystyle Let~ the~ locations~ be~ represented~ by~~~~~~~~(DCBA)^2 = hgfedcba$ \color\red {1041} $h=0~~So~~D=1...>~~~~~~~\because(4_{ Digit})^2 =7_ {Digit} \\ \sqrt{2,32,41,31}= 1524..>~~~~\therefore B=5~not~0.\\But~~ here~~ B=2,~with f=2;~so ~to~make~B=5~ change~ f~~from~2~to~7. \\ \sqrt{2,37,41,31} =1540\therefore~ B=4 ~and~ A=1.\\ \boxed{1541}$ \color\red {2221} $\displaystyle h=0~~So~~D=2...>~~~~~~\because (4_{ Digit})^2 =7_ {Digit} \\ \sqrt{2,20,11,21}= 1483..>\\But~here~D=1~so ~to~make~D=2~ change~ g~~from~2~to~7. .\\\sqrt{7,20,11,21}= 2683..~~~\\\therefore~ C=7....to~ make~ this~ possible~ f~ changes~ from ~~ 2~to~7. \\ \sqrt{7,70,11,21}= 2775..\\B=7~ and~A~can~not~be~5, ~ A=6.\\\boxed{2776}\\ \\B-A=1735_{10} =1230_{tech}\\~~ \\~~~Sorry ~my~ explanation~ requires~ improvement.$

Given the Technometic representations of 1041^2 = 2324131 and 2221^2 = 2201121:

There are many Technometic representations of 1041 and 2221:

1041 = 1041, 1591, 1541, 1596, 1546, 1091, 1041, 1096, 1046,... and any permissible arrangements according to the rule...

2221 = 2226, 2276, 2271, 2721, 2771, 2726, 2776, and any permissible arrangements according to the rule...

To find the only normal representation of 1041 and 2221 given their squares, we check first their highest digits and their remaining digits.

Firstly for 2221, if the normal representation has 7 as its highest digit, then it would exceed the number of digits required since (2abc)^2 < 10000000 < (7def)^2. (a,b,c,d,e, and f are digits)

So 2 is the highest digit (for the left-side) but (for the right-side) the highest digit is 2 in which no four-digit square (with 2 or 7 as the highest digit) satisfies this since 2000000 << (2abc)^2. Hence we have the digit 7 for the right-side.

From the strict inequality (2300)^2 < 7000000 < (2800)^2, we can conclude that 7 is the hundreds digit for the left-side (Thus, we have so far, 27bc and for the right-side, 7defghi).

Can you do bashing? It won't be so hard to bash the remaining 4 normal numbers 2721, 2726, 2771, 2776; squaring them, and finding the Technometric number. in which only 2776 satisfies the given Technometic equation.

Using the same techniques for 1041, we can bash for the remaining 4 normal numbers: 1541, 1546, 1591, 1596. Giving 1541 as the suitable normal number satisfying the Technometric equation.

Hence, A - B = 1235 = 1230 (in Technometic).