The largest 5-digit number that leaves a remainder of 5 when divided by both 8 and 9 is 99,941. To find the greatest number of 5 digits that will give a remainder of 5 when divided by 8 and 9 respectively we need to use the concept of LCM (Least Common Multiple, which is 72 in this case) to find a general form for the numbers that satisfy the given conditions (x = 72k + 5). Then, we will find the largest 5-digit number that fits this form.<\/p>\n\n\n\n
Complete Answer:<\/strong><\/p>\n\n\n\n Here is a step by step process to find the greatest number of 5 digits that will give a remainder of 5 when divided by 8 and 9 respectively<\/p>\n\n\n\n Step 1: Understanding the condition<\/strong><\/p>\n Let the required number be N <\/span>.<\/p>\n When N<\/span> is divided by 8, it leaves a remainder of 5. <\/span>N<\/span><\/span>\u2261<\/span><\/span><\/span>5<\/span>\u00a0<\/span>(<\/span>mod<\/span>\u00a0<\/span>8<\/span>)<\/span><\/p>\n This means <\/span>N<\/span><\/span>\u2212<\/span><\/span><\/span>5<\/span> is divisible by 8.<\/p>\n <\/li>\n Similarly, when <\/span>N<\/span> is divided by 9, it leaves a remainder of 5. <\/span>N<\/span><\/span>\u2261<\/span><\/span><\/span>5<\/span>\u00a0<\/span>(<\/span>mod<\/span>\u00a0<\/span>9<\/span>)<\/span><\/p>\n This means <\/span>N<\/span><\/span>\u2212<\/span><\/span><\/span>5<\/span> is divisible by 9.<\/p>\n <\/li>\n<\/ul>\n Hence, <\/span>N<\/span><\/span>\u2212<\/span><\/span><\/span>5<\/span> is a common multiple of 8 and 9.<\/p>\n \u00a0<\/p>\n Step 2: Finding the least common multiple (LCM) of 8 and 9<\/strong><\/p>\n The LCM of 8 and 9 can be calculated as:<\/p>\n <\/span>LCM<\/span>(<\/span>8<\/span>,<\/span><\/span>9<\/span>)<\/span><\/span>=<\/span><\/span><\/span>72<\/span><\/p>\n Thus, <\/span>N<\/span><\/span>\u2212<\/span><\/span><\/span>5<\/span> is a multiple of 72. \n \n \n N\n =\n 72\n k\n +\n 5\n \n N = 72k + 5\u00a0\n \n <\/span><\/span>N<\/span><\/span>=<\/span><\/span><\/span>72<\/span>k<\/span><\/span>+<\/span><\/span><\/span>5<\/span><\/p>\n \u00a0<\/p>\n Step 3: Finding the greatest 5-digit number<\/strong><\/p>\n The greatest 5-digit number is 99999. <\/span>72<\/span>k<\/span><\/span>\u2264<\/span><\/span><\/span>99994<\/span><\/p>\n Now divide both sides by 72 to find <\/span>k<\/span>:<\/p>\n <\/span>k<\/span><\/span>\u2264<\/span><\/span><\/span><\/span><\/span>72<\/span><\/span><\/span><\/span>99994<\/span>\u200b<\/span><\/span><\/span><\/span>\u2248<\/span><\/span><\/span>1388.805<\/span><\/p>\n Since <\/span>k<\/span> must be an integer, take the greatest integer less than or equal to 1388.805:<\/p>\n <\/span>k<\/span><\/span>=<\/span><\/span><\/span>1388<\/span><\/p>\n \u00a0<\/p>\n Step 4: Calculate <\/span><\/strong>N<\/strong><\/span><\/p>\n Substitute <\/span>k<\/span><\/span>=<\/span><\/span><\/span>1388<\/span> into <\/span>N<\/span><\/span>=<\/span><\/span><\/span>72<\/span>k<\/span><\/span>+<\/span><\/span><\/span>5<\/span>:<\/p>\n <\/span>N<\/span><\/span>=<\/span><\/span><\/span>72<\/span><\/span>\u00d7<\/span><\/span><\/span>1388<\/span><\/span>+<\/span><\/span><\/span>5<\/span><\/span>=<\/span><\/span><\/span>99936<\/span><\/span>+<\/span><\/span><\/span>5<\/span><\/span>=<\/span><\/span><\/span>99941<\/span><\/p>\n \u00a0<\/p>\n Step 5: Verify the solution<\/strong><\/p>\n Check if <\/span>99941<\/span> leaves a remainder of 5 when divided by 8:<\/p>\n <\/span>\n
Mathematically:<\/p>\n
Mathematically:<\/p>\n
Let <\/span>N<\/span><\/span>\u2212<\/span><\/span><\/span>5<\/span><\/span>=<\/span><\/span><\/span>72<\/span>k<\/span>, where <\/span>k<\/span> is a positive integer.
This gives:<\/p>\n
We want <\/span>N<\/span><\/span>=<\/span><\/span><\/span>72<\/span>k<\/span><\/span>+<\/span><\/span><\/span>5<\/span><\/span>\u2264<\/span><\/span><\/span>99999<\/span>.
Subtract 5 from both sides:<\/p>\n\n