Form 1015 2: Fill & Download for Free

This number is a bit larger than [math]16^N[/math], which is [math]4^N[/math] squared. So, if it’s a perfect square, it needs to be of the form [math](4^N [/math][math][/math][math]+ a)^2[/math], where [math]a[/math] is a positive integer.Now[math](4^N [/math][math][/math][math]+ a)^2 = 16^N [/math][math][/math][math]+ 2a \cdot 4^N [/math][math][/math][math]+ a^2[/math]So we need to find the [math]N[/math] for which[math]2^{50} [/math][math][/math][math]+ 2^{2030} = a \cdot 2^{2N+1} [/math][math][/math][math]+ a^2[/math]For some positive integer [math]a[/math]. The first observation is that if N>=1015, the RHS is bigger than the LHS, so we already have an initial upper bound for N. But we can do better than that.The LHS is equal to[math]2^{50} (2^{1980} [/math][math][/math][math]+ 1) [/math][math][/math][math][/math]So let’s try to look at the factoring of the RHS.[math]a \cdot 2^{2N+1} [/math][math][/math][math]+ a^2 = a \cdot (2^{2N+1} [/math][math][/math][math]+ a)[/math]which only has a shot of having 50 factors 2 if [math]a[/math] itself has a lot of 2’s in its factorization. To be precise, let [math]2^\alpha[/math] be the greatest power or 2 that divides [math]a[/math], i.e.[math]a = 2^{\alpha} \beta[/math], with [math]\beta[/math] odd.So[math]a \cdot (2^{2N+1} [/math][math][/math][math]+ a) = 2^{\alpha} \beta \cdot (2^{2N+1} [/math][math][/math][math]+ 2^{\alpha} \beta)[/math]We have two possibilities, either [math]\alpha < 2N [/math][math][/math][math]+ 1[/math], or not.If [math]\alpha < 2N [/math][math][/math][math]+ 1[/math], it equals[math]2^{2\alpha} \beta \cdot (2^{2N+1 - \alpha} [/math][math][/math][math]+ \beta)[/math]and that should be equal to[math]2^{50} (2^{1980} [/math][math][/math][math]+ 1) [/math][math][/math][math][/math]but all the factors except for [math]2^{2\alpha}[/math] are odd, so all the 2’s come from [math]2^{2\alpha}[/math], i.e.[math]2^{2\alpha} = 2^{50}[/math][math]\alpha = 25[/math]and[math]\beta \cdot (2^{2N-24} [/math][math][/math][math]+ \beta) = (2^{1980} [/math][math][/math][math]+ 1)[/math]We could go ahead and factorize [math](2^{1980} [/math][math][/math][math]+ 1)[/math], to exhaust all the posibilities, but we only want the greatest [math]N[/math], and clearly, the bigger [math]\beta[/math] is, the smaller [math]N[/math] is going to be.Checking then [math]\beta = 1[/math], we get[math]2N - 24 = 1980 [/math][math][/math][math][/math][math]N = 1002[/math]which is a solution.[math](4^{1002} [/math][math][/math][math]+ 2^{25})^2 = 16^{1002} [/math][math][/math][math]+ 2\cdot 2^{25} \cdot 4^{1002} [/math][math][/math][math]+ 2^{50} = 16^{1002} [/math][math][/math][math]+ 4^{1015} [/math][math][/math][math]+ 2^{50}[/math]If we assume it’s not, i.e. [math]\alpha >= 2N+1[/math].Then[math]2^{\alpha} \beta \cdot (2^{2N+1} [/math][math][/math][math]+ 2^{\alpha} \beta) = 2^{\alpha [/math][math][/math][math]+ 2N [/math][math][/math][math]+ 1} \beta \cdot (1 [/math][math][/math][math]+ 2^{\alpha - 2N - 1} \beta)[/math]And again, the first factor is the only even factor, and, by the same argument[math]\alpha [/math][math][/math][math]+ 2N [/math][math][/math][math]+ 1 = 50[/math]which means 2N < 50, and a fortiori, N is much less than 1002, which is already our best solution.So, 1002 is the largest N.

Depends totally on the digit [math]x[/math] is and the base you are using.If you use base [math]b[/math] ([math]b \in \mathbb N \land b \ge 8[/math]) and [math]x[/math] is the digit [math]y[/math] ([math]y \in \mathbb N \land 0 \le y \lt b[/math]) than you would have [math]10y7_{b}=(b^2+y)b+7[/math]. For [math]b=6 \lor 7[/math] the think looks like this:6 1005+2 1011_6 223 6 1015+2 1021_6 229 6 1025+2 1031_6 235 6 1035+2 1041_6 241 6 1045+2 1051_6 247 6 1055+2 1101_6 253 7 1005+2 1010_7 350 7 1015+2 1020_7 357 7 1025+2 1030_7 364 7 1035+2 1040_7 371 7 1045+2 1050_7 378 7 1055+2 1060_7 385 7 1065+2 1100_7 392 Ok enough messing with you:as others say consider the order of operations:Than [math]5 \times 10+2=50+2=52[/math].

Obviously you can just google the answer, but this is a fun one to actually figure out. Since I’m a MathCounts coach, I know it’s not 1001, because that’s a funky number — its prime factorization is 7*11*13. That also makes 1003 a really prime candidate to be the smallest 4-digit prime number, since it’s obviously not divisible by 2, 3, 5 (those divisibility rules are trivial) 7, 11, or 13 (because it’s close enough to 1001 that it can’t also be divisible by any of its prime factors). The square root of 1024 is 32, so we only have to check up to 31.17? 850+170 = 1020 — aw, dang! 1003 is divisible by 17.Okay, let’s try 1007. We know 2–17 are safe because it’s too close to 1001 and 1003. We only need to check 19, 23, 29, and 31.19: 950+57 = 1007. Aw, c’mon! It’s divisible by 19.Next up is 1009, which is close enough to 1001, 1003, and 1007 to be safe from 2–19 (well, it’s not divisible by 7 because 1008 is). Let’s try 23 . . .23: 1150–115 = 1035, minus 23 is 1012. Good!29: 870 + 145 = 1015. Good!31: 930 + 62 = 992, plus 31 is 1023. Good!So our winner is 1009. Did not expect that.