AIME 2026 I · 第 4 题

Solution 1

Let $x$ be equal to $a+b+ab$. Adding 1 to both sides, we get $a+b+ab+1=(a+1)(b+1)=x+1$. Because we know $a$ and $b$ have to be positive, this means that $x+1$ cannot be prime. We have $25$ primes less than $100$, but we have to also count $101$ since $x=100$ is still in the range of $1$ to $100$. Another thing we can notice is that perfect squares of primes don’t work either. The primes whose squares are less than $100$ are $2$,$3$,$5$, and $7$. Therefore, the answer is $100-26-4=\boxed{070}$.

- ChickensEatGrass

Solution 2 (Basically equivalent to solution 1, but a little more explanation)

Rewrite the expression as $a + b + ab = (a + 1)(b + 1)-1$.

Let $x = a + 1$ and $y = b + 1$. Then $x$ and $y$ are integers at least $2$, and since $a$ and $b$ are distinct, $x \neq y$. The expression becomes $xy - 1$, and the condition that the value be at most $100$ is equivalent to $xy \leq 101$.

Thus we are counting the integers of the form $xy-1$ with $x$ and $y$ distinct integers at least $2$ and product at most $101$. Different choices of $x$ and $y$ that give the same product produce the same value, so this reduces to counting integers $m = xy \neq 101$ that can be written as a product of two distinct integers at least $2$.

The number $m$ cannot be $1$ or a prime. Most composite numbers work, but the exception is when $m$ is a perfect square whose only factorization uses equal factors, since that would force $x = y$, which is not allowed. The only such squares at most $101$ are $4$, $9$, $25$, and $49$. Every other composite number at most $101$ has a factorization with two distinct factors at least $2$.

There are $101$ integers from $1$ to $101$. Removing $1$ leaves $100$. Among these, $26$ are prime, so there are $74$ composite numbers. Excluding the four squares $4$, $9$, $25$, and $49$ leaves $70$ valid values of $m$. Each gives a distinct integer $m - 1$ at most $100$.

Therefore, the number of integers less than or equal to $100$ that can be written in the specified form is $\boxed{070}$.

~Gray_Wolf

~ $\LaTeX$ by Logibyte

Solution 3 (Bash)

We take inspiration from the Sieve of Eratosthenes

Case 1: $a=1$ and $b>1 \rightarrow 1+2b$

We can do two things (and more generally for $a=2, 3,$ etc) 1. We find the smallest value that satisfies the expression $1+2b$, and use the period of $2$ to box numbers that are $2$ more than that minimum 2. Cross out any number below $1+2b$ that is not circled. Boxing $5$ and crossing out $1,2,3,4$ and circling all $5,7,9, \dots, 97, 99$ gives us the following:

\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \cancel{1} & \cancel{2} & \cancel{3} & \cancel{4} & \boxed{5} & 6 & \boxed{7} & 8 & \boxed{9} & 10 \\ \hline \boxed{11} & 12 & \boxed{13} & 14 & \boxed{15} & 16 & \boxed{17} & 18 & \boxed{19} & 20 \\ \hline \boxed{21} & 22 & \boxed{23} & 24 & \boxed{25} & 26 & \boxed{27} & 28 & \boxed{29} & 30 \\ \hline \boxed{31} & 32 & \boxed{33} & 34 & \boxed{35} & 36 & \boxed{37} & 38 & \boxed{39} & 40 \\ \hline \boxed{41} & 42 & \boxed{43} & 44 & \boxed{45} & 46 & \boxed{47} & 48 & \boxed{49} & 50 \\ \hline \boxed{51} & 52 & \boxed{53} & 54 & \boxed{55} & 56 & \boxed{57} & 58 & \boxed{59} & 60 \\ \hline \boxed{61} & 62 & \boxed{63} & 64 & \boxed{65} & 66 & \boxed{67} & 68 & \boxed{69} & 70 \\ \hline \boxed{71} & 72 & \boxed{73} & 74 & \boxed{75} & 76 & \boxed{77} & 78 & \boxed{79} & 80 \\ \hline \boxed{81} & 82 & \boxed{83} & 84 & \boxed{85} & 86 & \boxed{87} & 88 & \boxed{89} & 90 \\ \hline \boxed{91} & 92 & \boxed{93} & 94 & \boxed{95} & 96 & \boxed{97} & 98 & \boxed{99} & 100 \\ \hline \end{array}

Case 2: we have $a=2, b>2 \rightarrow 2+3b$

1. minimum is $b=3 \rightarrow 11.$

2. we cross out $6, 8, 10.$

3 we add $14, 20, 26 \dots$ etc.

\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \cancel{1} & \cancel{2} & \cancel{3} & \cancel{4} & \boxed{5} & \cancel{6} & \boxed{7} & \cancel{8} & \boxed{9} & \cancel{10} \\ \hline \boxed{11} & 12 & \boxed{13} & \boxed{14} & \boxed{15} & 16 & \boxed{17} & 18 & \boxed{19} & \boxed{20} \\ \hline \boxed{21} & 22 & \boxed{23} & 24 & \boxed{25} & \boxed{26} & \boxed{27} & 28 & \boxed{29} & 30 \\ \hline \boxed{31} & \boxed{32} & \boxed{33} & 34 & \boxed{35} & 36 & \boxed{37} & \boxed{38} & \boxed{39} & 40 \\ \hline \boxed{41} & 42 & \boxed{43} & \boxed{44} & \boxed{45} & 46 & \boxed{47} & 48 & \boxed{49} & \boxed{50} \\ \hline \boxed{51} & 52 & \boxed{53} & 54 & \boxed{55} & \boxed{56} & \boxed{57} & 58 & \boxed{59} & 60 \\ \hline \boxed{61} & \boxed{62} & \boxed{63} & 64 & \boxed{65} & 66 & \boxed{67} & \boxed{68} & \boxed{69} & 70 \\ \hline \boxed{71} & 72 & \boxed{73} & \boxed{74} & \boxed{75} & 76 & \boxed{77} & 78 & \boxed{79} & \boxed{80} \\ \hline \boxed{81} & 82 & \boxed{83} & 84 & \boxed{85} & \boxed{86} & \boxed{87} & 88 & \boxed{89} & 90 \\ \hline \boxed{91} & \boxed{92} & \boxed{93} & 94 & \boxed{95} & 96 & \boxed{97} & \boxed{98} & \boxed{99} & 100 \\ \hline \end{array}

Case 3: $3+4b.$

1. minimum is $19.$

2. note nothing extra gets boxed (odd).

\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \cancel{1} & \cancel{2} & \cancel{3} & \cancel{4} & \boxed{5} & \cancel{6} & \boxed{7} & \cancel{8} & \boxed{9} & \cancel{10} \\ \hline \boxed{11} & 12 & \boxed{13} & \boxed{14} & \boxed{15} & \cancel{16} & \boxed{17} & \cancel{18} & \boxed{19} & \boxed{20} \\ \hline \boxed{21} & 22 & \boxed{23} & 24 & \boxed{25} & \boxed{26} & \boxed{27} & 28 & \boxed{29} & 30 \\ \hline \boxed{31} & \boxed{32} & \boxed{33} & 34 & \boxed{35} & 36 & \boxed{37} & \boxed{38} & \boxed{39} & 40 \\ \hline \boxed{41} & 42 & \boxed{43} & \boxed{44} & \boxed{45} & 46 & \boxed{47} & 48 & \boxed{49} & \boxed{50} \\ \hline \boxed{51} & 52 & \boxed{53} & 54 & \boxed{55} & \boxed{56} & \boxed{57} & 58 & \boxed{59} & 60 \\ \hline \boxed{61} & \boxed{62} & \boxed{63} & 64 & \boxed{65} & 66 & \boxed{67} & \boxed{68} & \boxed{69} & 70 \\ \hline \boxed{71} & 72 & \boxed{73} & \boxed{74} & \boxed{75} & 76 & \boxed{77} & 78 & \boxed{79} & \boxed{80} \\ \hline \boxed{81} & 82 & \boxed{83} & 84 & \boxed{85} & \boxed{86} & \boxed{87} & 88 & \boxed{89} & 90 \\ \hline \boxed{91} & \boxed{92} & \boxed{93} & 94 & \boxed{95} & 96 & \boxed{97} & \boxed{98} & \boxed{99} & 100 \\ \hline \end{array}

Case 4: $4+5b.$

1. minimum is $29.$

2. column of $4$ gets boxed.

\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \cancel{1} & \cancel{2} & \cancel{3} & \cancel{4} & \boxed{5} & \cancel{6} & \boxed{7} & \cancel{8} & \boxed{9} & \cancel{10} \\ \hline \boxed{11} & \cancel{12} & \boxed{13} & \boxed{14} & \boxed{15} & \cancel{16} & \boxed{17} & \cancel{18} & \boxed{19} & \boxed{20} \\ \hline \boxed{21} & \cancel{22} & \boxed{23} & \cancel{24} & \boxed{25} & \boxed{26} & \boxed{27} & \cancel{28} & \boxed{29} & 30 \\ \hline \boxed{31} & \boxed{32} & \boxed{33} & \boxed{34} & \boxed{35} & 36 & \boxed{37} & \boxed{38} & \boxed{39} & 40 \\ \hline \boxed{41} & 42 & \boxed{43} & \boxed{44} & \boxed{45} & 46 & \boxed{47} & 48 & \boxed{49} & \boxed{50} \\ \hline \boxed{51} & 52 & \boxed{53} & \boxed{54} & \boxed{55} & \boxed{56} & \boxed{57} & 58 & \boxed{59} & 60 \\ \hline \boxed{61} & \boxed{62} & \boxed{63} & \boxed{64} & \boxed{65} & 66 & \boxed{67} & \boxed{68} & \boxed{69} & 70 \\ \hline \boxed{71} & 72 & \boxed{73} & \boxed{74} & \boxed{75} & 76 & \boxed{77} & 78 & \boxed{79} & \boxed{80} \\ \hline \boxed{81} & 82 & \boxed{83} & \boxed{84} & \boxed{85} & \boxed{86} & \boxed{87} & 88 & \boxed{89} & 90 \\ \hline \boxed{91} & \boxed{92} & \boxed{93} & \boxed{94} & \boxed{95} & 96 & \boxed{97} & \boxed{98} & \boxed{99} & 100 \\ \hline \end{array}

Case 5: $5+6b.$

1. minimum is $41.$

2. no boxes (odd).

\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \cancel{1} & \cancel{2} & \cancel{3} & \cancel{4} & \boxed{5} & \cancel{6} & \boxed{7} & \cancel{8} & \boxed{9} & \cancel{10} \\ \hline \boxed{11} & \cancel{12} & \boxed{13} & \boxed{14} & \boxed{15} & \cancel{16} & \boxed{17} & \cancel{18} & \boxed{19} & \boxed{20} \\ \hline \boxed{21} & \cancel{22} & \boxed{23} & \cancel{24} & \boxed{25} & \boxed{26} & \boxed{27} & \cancel{28} & \boxed{29} & \cancel{30} \\ \hline \boxed{31} & \boxed{32} & \boxed{33} & \boxed{34} & \boxed{35} & \cancel{36} & \boxed{37} & \boxed{38} & \boxed{39} & \cancel{40} \\ \hline \boxed{41} & 42 & \boxed{43} & \boxed{44} & \boxed{45} & 46 & \boxed{47} & 48 & \boxed{49} & \boxed{50} \\ \hline \boxed{51} & 52 & \boxed{53} & \boxed{54} & \boxed{55} & \boxed{56} & \boxed{57} & 58 & \boxed{59} & 60 \\ \hline \boxed{61} & \boxed{62} & \boxed{63} & \boxed{64} & \boxed{65} & 66 & \boxed{67} & \boxed{68} & \boxed{69} & 70 \\ \hline \boxed{71} & 72 & \boxed{73} & \boxed{74} & \boxed{75} & 76 & \boxed{77} & 78 & \boxed{79} & \boxed{80} \\ \hline \boxed{81} & 82 & \boxed{83} & \boxed{84} & \boxed{85} & \boxed{86} & \boxed{87} & 88 & \boxed{89} & 90 \\ \hline \boxed{91} & \boxed{92} & \boxed{93} & \boxed{94} & \boxed{95} & 96 & \boxed{97} & \boxed{98} & \boxed{99} & 100 \\ \hline \end{array}

Case 6: $6+7b.$

1. minimum is $55.$

2. no trick here (it's a bash what do you expect).

\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \cancel{1} & \cancel{2} & \cancel{3} & \cancel{4} & \boxed{5} & \cancel{6} & \boxed{7} & \cancel{8} & \boxed{9} & \cancel{10} \\ \hline \boxed{11} & \cancel{12} & \boxed{13} & \boxed{14} & \boxed{15} & \cancel{16} & \boxed{17} & \cancel{18} & \boxed{19} & \boxed{20} \\ \hline \boxed{21} & \cancel{22} & \boxed{23} & \cancel{24} & \boxed{25} & \boxed{26} & \boxed{27} & \cancel{28} & \boxed{29} & \cancel{30} \\ \hline \boxed{31} & \boxed{32} & \boxed{33} & \boxed{34} & \boxed{35} & \cancel{36} & \boxed{37} & \boxed{38} & \boxed{39} & \cancel{40} \\ \hline \boxed{41} & \cancel{42} & \boxed{43} & \boxed{44} & \boxed{45} & \cancel{46} & \boxed{47} & \cancel{48} & \boxed{49} & \boxed{50} \\ \hline \boxed{51} & \cancel{52} & \boxed{53} & \boxed{54} & \boxed{55} & \boxed{56} & \boxed{57} & 58 & \boxed{59} & 60 \\ \hline \boxed{61} & \boxed{62} & \boxed{63} & \boxed{64} & \boxed{65} & 66 & \boxed{67} & \boxed{68} & \boxed{69} & 70 \\ \hline \boxed{71} & 72 & \boxed{73} & \boxed{74} & \boxed{75} & \boxed{76} & \boxed{77} & 78 & \boxed{79} & \boxed{80} \\ \hline \boxed{81} & 82 & \boxed{83} & \boxed{84} & \boxed{85} & \boxed{86} & \boxed{87} & 88 & \boxed{89} & \boxed{90} \\ \hline \boxed{91} & \boxed{92} & \boxed{93} & \boxed{94} & \boxed{95} & 96 & \boxed{97} & \boxed{98} & \boxed{99} & 100 \\ \hline \end{array}

Case 7: $7+8b.$

1. minimum is $71.$

2. no box (odd).

\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \cancel{1} & \cancel{2} & \cancel{3} & \cancel{4} & \boxed{5} & \cancel{6} & \boxed{7} & \cancel{8} & \boxed{9} & \cancel{10} \\ \hline \boxed{11} & \cancel{12} & \boxed{13} & \boxed{14} & \boxed{15} & \cancel{16} & \boxed{17} & \cancel{18} & \boxed{19} & \boxed{20} \\ \hline \boxed{21} & \cancel{22} & \boxed{23} & \cancel{24} & \boxed{25} & \boxed{26} & \boxed{27} & \cancel{28} & \boxed{29} & \cancel{30} \\ \hline \boxed{31} & \boxed{32} & \boxed{33} & \boxed{34} & \boxed{35} & \cancel{36} & \boxed{37} & \boxed{38} & \boxed{39} & \cancel{40} \\ \hline \boxed{41} & \cancel{42} & \boxed{43} & \boxed{44} & \boxed{45} & \cancel{46} & \boxed{47} & \cancel{48} & \boxed{49} & \boxed{50} \\ \hline \boxed{51} & \cancel{52} & \boxed{53} & \boxed{54} & \boxed{55} & \boxed{56} & \boxed{57} & \cancel{58} & \boxed{59} & \cancel{60} \\ \hline \boxed{61} & \boxed{62} & \boxed{63} & \boxed{64} & \boxed{65} & \cancel{66} & \boxed{67} & \boxed{68} & \boxed{69} & \cancel{70} \\ \hline \boxed{71} & 72 & \boxed{73} & \boxed{74} & \boxed{75} & \boxed{76} & \boxed{77} & 78 & \boxed{79} & \boxed{80} \\ \hline \boxed{81} & 82 & \boxed{83} & \boxed{84} & \boxed{85} & \boxed{86} & \boxed{87} & 88 & \boxed{89} & \boxed{90} \\ \hline \boxed{91} & \boxed{92} & \boxed{93} & \boxed{94} & \boxed{95} & 96 & \boxed{97} & \boxed{98} & \boxed{99} & 100 \\ \hline \end{array}

Case 8: $8+9b.$

1. minimum $89.$

2. (case 9 exceeds 100) so we can cross everything out now.

\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \cancel{1} & \cancel{2} & \cancel{3} & \cancel{4} & \boxed{5} & \cancel{6} & \boxed{7} & \cancel{8} & \boxed{9} & \cancel{10} \\ \hline \boxed{11} & \cancel{12} & \boxed{13} & \boxed{14} & \boxed{15} & \cancel{16} & \boxed{17} & \cancel{18} & \boxed{19} & \boxed{20} \\ \hline \boxed{21} & \cancel{22} & \boxed{23} & \cancel{24} & \boxed{25} & \boxed{26} & \boxed{27} & \cancel{28} & \boxed{29} & \cancel{30} \\ \hline \boxed{31} & \boxed{32} & \boxed{33} & \boxed{34} & \boxed{35} & \cancel{36} & \boxed{37} & \boxed{38} & \boxed{39} & \cancel{40} \\ \hline \boxed{41} & \cancel{42} & \boxed{43} & \boxed{44} & \boxed{45} & \cancel{46} & \boxed{47} & \cancel{48} & \boxed{49} & \boxed{50} \\ \hline \boxed{51} & \cancel{52} & \boxed{53} & \boxed{54} & \boxed{55} & \boxed{56} & \boxed{57} & \cancel{58} & \boxed{59} & \cancel{60} \\ \hline \boxed{61} & \boxed{62} & \boxed{63} & \boxed{64} & \boxed{65} & \cancel{66} & \boxed{67} & \boxed{68} & \boxed{69} & \cancel{70} \\ \hline \boxed{71} & \cancel{72} & \boxed{73} & \boxed{74} & \boxed{75} & \boxed{76} & \boxed{77} & \cancel{78} & \boxed{79} & \boxed{80} \\ \hline \boxed{81} & \cancel{82} & \boxed{83} & \boxed{84} & \boxed{85} & \boxed{86} & \boxed{87} & \cancel{88} & \boxed{89} & \boxed{90} \\ \hline \boxed{91} & \boxed{92} & \boxed{93} & \boxed{94} & \boxed{95} & \cancel{96} & \boxed{97} & \boxed{98} & \boxed{99} & \cancel{100} \\ \hline \end{array}

Counting the ones that are boxed we get $\boxed{070}$ numbers.

~Soccerstar9

~arushkrisp

Solution 4 (Simon's favorite factoring trick!)

Using Simon's favorite factoring trick, $a+b+ab+1\leq 101$, or $(a+1)(b+1)\leq 101$. Because $a, b$ are distinct, all squares that don't have a non square representation, which at or below $100$ are removed from the set, which are only five of them $(1^2, 2^2, 3^2, 5^2, 7^2)$. This means $95$ remain.

More can be removed, such as primes, that cannot be factorized into a number less than or equal to $101$. By beast academy memorization, there are 25 primes less than or equal to 101. $95 - 25 = \fbox{70}$, which is the answer.

~minor $\LaTeX$ formatting by yvz2900 (who did not write the original solution)

Video Solution (Fast and Easy 🔥🚀)

2026 AIME I #4

By piacademyus.org

Video Solution (Logic)

https://www.youtube.com/watch?v=iCKaduIZu5Q - Continuum Math