可知其中僅c=1時，同一列中會有兩數之差恰為c=1的情況：3×5+7×

可知其中僅c=1時，同一列中會有兩數之差恰為c=1的情況：3×5+7×9≡1與3*7+5*9≡0故可判斷ab 、de為39與57，或是75與93(5分)。分別代回驗算，可知75×93-1=6974的四個數碼不相同，故不合；而57×39-1=2222，滿足題意。因此可能的五位數有57139、39157，其最大值為57139。(5分)

ผลลัพธ์ (อังกฤษ) 1:

Only when the c=1, there is difference between the two numbers in the same column as c=1: 3*7 5*9 and 3x5 7x9 ≡ 1 ≡ 0 you can determine AB, de 39 and 57 or 75 to 93 (5 points). Checking, respectively, four digital 75x93-1=6974 is not the same, it is not, and 57x39-1=2222 to meet the meaning. There may therefore be five digits 57139, 39157, and has a maximum value of 57139. (5 points)

ผลลัพธ์ (อังกฤษ) 2:

Found that c = 1 in which only the difference of two numbers in the same column have exactly the case c = 1: 3 × 5 +7 × 9 ≡ 1 and 3 * 7 +5 * 9 ≡ 0 and therefore can determine ab, de 39 and 57, or 75 and 93 (5 points). Substituting checking back, we can see 75 × 93-1 = 6974 four digital is not the same, so substandard; while 57 × 39-1 = 2222, meet Italy. Therefore, the number of possible five have 57139,39157, its maximum value is 57139. (5 points)

ผลลัพธ์ (อังกฤษ) 3:

May know when only c=1, in the same row will have the difference of two number exactly is the c=1 situation: 3×5 7×9≡1 and 3*7 5*9≡0, therefore may judge ab and de is 39 and 57, or 75 and 93 (5 points). The bringing back checking calculation, may know 75×93-1=6974 four numerical codes not to be the same separately, therefore does not gather; But 57×39-1=2222, satisfies topic Italy. Therefore the possible five figures have 57139 and 39157, its maximum value is 57139. (5 points)